Miscellaneous Exercise Integrals Solution - NCERT Class 12


Chapter 7 Ex.7.ME Question 1

Integrate \(\frac{1}{{x - {x^3}}}\)

 

Solution

 

\(\frac{1}{{x - {x^3}}} = \frac{1}{{x\left( {1 - {x^2}} \right)}} = \frac{1}{{x\left( {1 - x} \right)\left( {1 + x} \right)}}\)

Let \(\frac{1}{{x\left( {1 - x} \right)\left( {1 + x} \right)}} = \frac{A}{x} + \frac{B}{{\left( {1 - x} \right)}} + \frac{C}{{\left( {1 + x} \right)}}\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\)

\[\begin{align} &\Rightarrow \; 1 = A\left( {1 - {x^2}} \right) + Bx\left( {1 + x} \right) + Cx\left( {1 - x} \right)\\& \Rightarrow \; 1 = A - A{x^2} + Bx + B{x^2} + Cx - C{x^2}\end{align}\]

Equating the coefficients of \({x^2},x\)and constant terms, we get

\[\begin{align} - A + B - C &= 0\\B + C &= 0\\A &= 1\end{align}\]

On solving these equations, we get

\[\begin{align}A = 1\\B = \frac{1}{2}\\C = - \frac{1}{2}\end{align}\]

From equation (\(1\)), we get

\[\begin{align}\frac{1}{{x\left( {1 - x} \right)\left( {1 + x} \right)}}& = \frac{1}{x} + \frac{1}{{2\left( {1 - x} \right)}} - \frac{1}{{2\left( {1 + x} \right)}}\\& \Rightarrow \; \int {\frac{1}{{x\left( {1 - x} \right)\left( {1 + x} \right)}}} dx = \int {\frac{1}{x}dx} + \frac{1}{2}\int {\frac{1}{{\left( {1 - x} \right)}}dx} - \frac{1}{2}\int {\frac{1}{{\left( {1 + x} \right)}}} dx\\ &= \log \left| x \right| - \frac{1}{2}\log \left| {\left( {1 - x} \right)} \right| - \frac{1}{2}\log \left| {\left( {1 + x} \right)} \right|\\ &= \log \left| x \right| - \log \left| {{{\left( {1 - x} \right)}^{\frac{1}{2}}}} \right| - \log \left| {{{\left( {1 + x} \right)}^{\frac{1}{2}}}} \right| = \log \left| {\frac{x}{{{{\left( {1 - x} \right)}^{\frac{1}{2}}}{{\left( {1 + x} \right)}^{\frac{1}{2}}}}}} \right| + C\\& = \log \left| {{{\left( {\frac{{{x^2}}}{{1 - {x^2}}}} \right)}^{\frac{1}{2}}}} \right| + C = \frac{1}{2}\log \left| {\frac{{{x^2}}}{{1 - {x^2}}}} \right| + C\end{align}\]

Chapter 7 Ex.7.ME Question 2

Integrate \(\frac{1}{{\sqrt {x + a} + \sqrt {x + b} }}\)

 

Solution

 

\[\begin{align}\frac{1}{{\sqrt {x + a} + \sqrt {x + b} }} &= \frac{1}{{\sqrt {x + a} + \sqrt {x + b} }} \times \frac{{\sqrt {x + a} - \sqrt {x + b} }}{{\sqrt {x + a} - \sqrt {x + b} }}\\ &= \frac{{\sqrt {x + a} - \sqrt {x + b} }}{{\left( {x + a} \right) - \left( {x + b} \right)}} = \frac{{\left( {\sqrt {x + a} - \sqrt {x + b} } \right)}}{{a - b}}\\& \Rightarrow \; \int {\frac{1}{{\sqrt {x + a} + \sqrt {x + b} }}dx} = \frac{1}{{a - b}}\int {\left( {\sqrt {x + a} - \sqrt {x + b} } \right)dx} \\& = \frac{1}{{\left( {a - b} \right)}}\left[ {\frac{{{{\left( {x + a} \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}} - \frac{{{{\left( {x + b} \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right] = \frac{2}{{3\left( {a - b} \right)}}\left[ {{{\left( {x + a} \right)}^{\frac{3}{2}}} - {{\left( {x + b} \right)}^{\frac{3}{2}}}} \right] + C\end{align}\]

Chapter 7 Ex.7.ME Question 3

Integrate \(\frac{1}{{x\sqrt {ax - {x^2}} }}\,\,\,\,\,\,\,\,\left[ {{\rm{Hint: }}x = \frac{a}{t}} \right]\)

 

Solution

 

\(\frac{1}{{x\sqrt {ax - {x^2}} }}\)

Let \(x = \frac{a}{t} \Rightarrow \; dx = - \frac{a}{{{t^2}}}dt\)

\[\begin{align}& \Rightarrow \; \int {\frac{1}{{x\sqrt {ax - {x^2}} }}dx = \int {\frac{1}{{\frac{a}{t}\sqrt {a.\frac{a}{t} - {{\left( {\frac{a}{t}} \right)}^2}} }}} } \left( { - \frac{a}{{{t^2}}}dt} \right)\\ &= - \int {\frac{1}{{at}}} \frac{1}{{\sqrt {\frac{1}{t} - \frac{1}{{{t^2}}}} }}dt = - \frac{1}{a}\int {\frac{1}{{\sqrt {\frac{{{t^2}}}{t} - \frac{{{t^2}}}{{{t^2}}}} }}} dt\\ &= - \frac{1}{a}\int {\frac{1}{{\sqrt {t - 1} }}} dt\\ &= - \frac{1}{a}\left[ {2\sqrt {t - 1} } \right] + C\\& = - \frac{1}{a}\left[ {2\sqrt {\frac{a}{x} - 1} } \right] + C\\& = - \frac{2}{a}\left( {\sqrt {\frac{{a - x}}{x}} } \right) + C\end{align}\]

Chapter 7 Ex.7.ME Question 4

Integrate \(\frac{1}{{{x^2}{{\left( {{x^4} + 1} \right)}^{\frac{3}{4}}}}}\)

 

Solution

 

\(\frac{1}{{{x^2}{{\left( {{x^4} + 1} \right)}^{\frac{3}{4}}}}}\)

Multiplying and dividing by \({x^{ - 3}}\), we get

\[\begin{align}\frac{{{x^{ - 3}}}}{{{x^2}{x^{ - 3}}{{\left( {{x^4} + 1} \right)}^{\frac{3}{4}}}}} &= \frac{{{x^{ - 3}}{{\left( {{x^4} + 1} \right)}^{\frac{{ - 3}}{4}}}}}{{{x^2}{x^{ - 3}}}}\\ &= \frac{{{{\left( {{x^4} + 1} \right)}^{\frac{{ - 3}}{4}}}}}{{{x^5}{{\left( {{x^4}} \right)}^{ - \frac{3}{4}}}}} = \frac{1}{{{x^5}}}{\left( {\frac{{{x^4} + 1}}{{{x^4}}}} \right)^{ - \frac{3}{4}}}\\& = \frac{1}{{{x^5}}}{\left( {1 + \frac{1}{{{x^4}}}} \right)^{ - \frac{3}{4}}}\end{align}\]

Let

\[\begin{align}\frac{1}{{{x^4}}} &= t \Rightarrow \; - \frac{4}{{{x^5}}}dx = dt \Rightarrow \; \frac{1}{{{x^5}}}dx = - \frac{{dt}}{4}\\&\therefore \int {\frac{1}{{{x^2}{{\left( {{x^4} + 1} \right)}^{\frac{3}{4}}}}}dx = \int {\frac{1}{{{x^5}}}} {{\left( {1 + \frac{1}{{{x^4}}}} \right)}^{ - \frac{3}{4}}}dx} = - \frac{1}{4}\int {{{\left( {1 + t} \right)}^{ - \frac{3}{4}}}} dt\\&= - \frac{1}{4}\left[ {\frac{{{{\left( {1 + t} \right)}^{\frac{1}{4}}}}}{{\frac{1}{4}}}} \right] + C = - \frac{1}{4}\frac{{{{\left( {1 + \frac{1}{{{x^4}}}} \right)}^{\frac{1}{4}}}}}{{\frac{1}{4}}} + C\\& = - {\left( {1 + \frac{1}{{{x^4}}}} \right)^{\frac{1}{4}}} + C\end{align}\]

Chapter 7 Ex.7.ME Question 5

Integrate \(\frac{1}{{{x^{\frac{1}{2}}} + {x^{\frac{1}{3}}}}}\,\,\,\,\,\,\left[ {{\rm{Hint: }}\frac{1}{{{x^{\frac{1}{2}}} + {x^{\frac{1}{3}}}}} = \frac{1}{{{x^{\frac{1}{3}}}\left( {1 + {x^{\frac{1}{6}}}} \right)}}{\rm{ put }}x = {t^6}} \right]\)

 

Solution

 

\(\frac{1}{{{x^{\frac{1}{2}}} + {x^{\frac{1}{3}}}}} = \frac{1}{{{x^{\frac{1}{3}}}\left( {1 + {x^{\frac{1}{6}}}} \right)}}\)

Let \(x = {t^6} \Rightarrow \; dx = 6{t^5}dt\)

\[\begin{align}&\therefore \int {\frac{1}{{{x^{\frac{1}{2}}} + {x^{\frac{1}{3}}}}}dx = \int {\frac{1}{{{x^{\frac{1}{3}}}\left( {1 + {x^{\frac{1}{6}}}} \right)}}} } dx = \int {\frac{{6{t^5}}}{{{t^2}\left( {1 + t} \right)}}} dt\\& = 6\int {\frac{{{t^3}}}{{\left( {1 + t} \right)}}} dt\end{align}\]

\[\begin{align}{\text{Adding and Substracting 1 in Numerator}} &= 6\int {\frac{{{t^3} + 1 - 1}}{{1 + t}}} dt\\&= 6\int {\left( {\frac{{{t^3} + 1}}{{1 + t}} - \frac{1}{{1 + t}}} \right)} dt\\{\text{Using }}{a^3} + {b^3}& = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\\&= 6\int {\left\{ {\frac{{\left( {1 + t} \right)\left( {{t^2} + {1^2} - 1 \times t} \right)}}{{1 + t}} - \frac{1}{{1 + t}}} \right\}} dt\\&= 6\int {\left\{ {\left( {{t^2} - t + 1} \right) - \frac{1}{{1 + t}}} \right\}} dt\\&= 6\left[ {\left( {\frac{{{t^3}}}{3}} \right) - \left( {\frac{{{t^2}}}{2}} \right) + t - \log \left| {1 + t} \right|} \right]\\&= 2{x^{\frac{1}{2}}} - 3{x^{\frac{1}{3}}} + 6{x^{\frac{1}{6}}} - 6\log \left( {1 + {x^{\frac{1}{6}}}} \right) + C\\& = 2\sqrt x - 3{x^{\frac{1}{3}}} + 6{x^{\frac{1}{6}}} - 6\log \left( {1 + {x^{\frac{1}{6}}}} \right) + C\end{align}\]

Chapter 7 Ex.7.ME Question 6

Integrate \(\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} + 9} \right)}}\)

 

Solution

 

Consider, \(\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} + 9} \right)}} = \frac{A}{{\left( {x + 1} \right)}} + \frac{{Bx + C}}{{\left( {{x^2} + 9} \right)}}\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\)

\[\begin{align}& \Rightarrow \; 5x = A\left( {{x^2} + 9} \right) + \left( {Bx + C} \right)\left( {x + 1} \right)\\ &\Rightarrow \; 5x = A{x^2} + 9A + B{x^2} + Bx + Cx + C\end{align}\]

Equating the coefficients of \({x^2},x\)and constant term, we get

\[\begin{align}A + B &= 0\\B + C &= 5\\9A + C &= 0\end{align}\]

On solving these equations, we get

\[\begin{align}A &= - \frac{1}{2}\\B &= \frac{1}{2}\\C &= \frac{9}{2}\end{align}\]

From equation (\(1\)), we get

\[\begin{align}\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} + 9} \right)}} &= \frac{{ - 1}}{{2\left( {x + 1} \right)}} + \frac{{\frac{x}{2} + \frac{9}{2}}}{{\left( {{x^2} + 9} \right)}}\\\int {\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} + 9} \right)}}} dx &= \int {\left\{ {\frac{{ - 1}}{{2\left( {x + 1} \right)}} + \frac{{\left( {x + 9} \right)}}{{2\left( {{x^2} + 9} \right)}}} \right\}} dx\\& = - \frac{1}{2}\log \left| {x + 1} \right| + \frac{1}{2}\int {\frac{x}{{{x^2} + 9}}} dx + \frac{9}{2}\int {\frac{1}{{{x^2} + 9}}} dx \\&= - \frac{1}{2}\log \left| {x + 1} \right| + \frac{1}{4}\int {\frac{{2x}}{{{x^2} + 9}}} dx + \frac{9}{2}\int {\frac{1}{{{x^2} + 9}}} dx\\ &= - \frac{1}{2}\log \left| {x + 1} \right| + \frac{1}{4}\log \left| {{x^2} + 9} \right| + \frac{9}{2}.\frac{1}{3}{\tan ^{ - 1}}\frac{x}{3} + C\\& = - \frac{1}{2}\log \left| {x + 1} \right| + \frac{1}{4}\log \left( {{x^2} + 9} \right) + \frac{3}{2}{\tan ^{ - 1}}\frac{x}{3} + C\end{align}\]

Chapter 7 Ex.7.ME Question 7

Integrate \(\frac{{\sin x}}{{\sin \left( {x - a} \right)}}\)

 

Solution

 

\(\frac{{\sin x}}{{\sin \left( {x - a} \right)}}\)

Put, \(x - a = t \Rightarrow \; dx = dt\)

\[\begin{align}\int {\frac{{\sin x}}{{\sin \left( {x - a} \right)}}dx} &= \int {\frac{{\sin \left( {t + a} \right)}}{{\sin t}}} dt\\ &= \int {\frac{{\sin t\cos a + \cos t\sin a}}{{\sin t}}} dt = \int {\left( {\cos a + \cot t\sin a} \right)} dt\\ &= t\cos a + \sin a\log \left| {\sin t} \right| + {C_1}\\ &= \left( {x - a} \right)\cos a + \sin a\log \left| {\sin \left( {x - a} \right)} \right| + {C_1}\\ &= x\cos a + \sin a\log \left| {\sin \left( {x - a} \right)} \right| - a\cos a + {C_1}\\ &= \sin a\log \left| {\sin \left( {x - a} \right)} \right| + x\cos a + C\end{align}\]

Chapter 7 Ex.7.ME Question 8

Integrate \(\frac{{{e^{5\log x}} - {e^{4\log x}}}}{{{e^{3\log x}} - {e^{2\log x}}}}\)

 

Solution

 

\[\begin{align}\frac{{{e^{5\log x}} - {e^{4\log x}}}}{{{e^{3\log x}} - {e^{2\log x}}}}& = \frac{{{e^{4\log x}}\left( {{e^{\log x}} - 1} \right)}}{{{e^{2\log x}}\left( {{e^{\log x}} - 1} \right)}}\\ &= {e^{2\log x}}\\ &= {e^{\log {x^2}}}\\ &= {x^2}\\&\therefore \int {\frac{{{e^{5\log x}} - {e^{4\log x}}}}{{{e^{3\log x}} - {e^{2\log x}}}}dx} = \int {{x^2}} dx = \frac{{{x^3}}}{3} + C\end{align}\]

Chapter 7 Ex.7.ME Question 9

Integrate \(\frac{{\cos x}}{{\sqrt {4 - {{\sin }^2}x} }}\)

 

Solution

 

\(\frac{{\cos x}}{{\sqrt {4 - {{\sin }^2}x} }}\)

Put, \(\sin x = t \Rightarrow \; \cos xdx = dt\)

\[\begin{align} &\Rightarrow \; \int {\frac{{\cos x}}{{\sqrt {4 - {{\sin }^2}x} }}} dx = \int {\frac{{dt}}{{\sqrt {{{\left( 2 \right)}^2} - {{\left( t \right)}^2}} }}} \\& = {\sin ^{ - 1}}\left( {\frac{t}{2}} \right) + C\\& = {\sin ^{ - 1}}\left( {\frac{{\sin x}}{2}} \right) + C\\ &= \frac{x}{2} + C\end{align}\]

Chapter 7 Ex.7.ME Question 10

Integrate \(\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}\)

 

Solution

 

\[\begin{align}\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}& = \frac{{\left( {{{\sin }^4}x - {{\cos }^4}x} \right)\left( {{{\sin }^4}x + {{\cos }^4}x} \right)}}{{{{\sin }^2}x + {{\cos }^2}x - {{\sin }^2}x{{\cos }^2}x - {{\sin }^2}x{{\cos }^2}x}}\\ &= \frac{{\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{\left( {{{\sin }^2}x - {{\sin }^2}x{{\cos }^2}x} \right) + \left( {{{\cos }^2}x - {{\sin }^2}x{{\cos }^2}x} \right)}}\\&= \frac{{\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right)}}{{{{\sin }^2}x\left( {1 - {{\cos }^2}x} \right) + {{\cos }^2}x\left( {1 - {{\sin }^2}x} \right)}}\\ &= \frac{{ - \left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\cos }^2}x - {{\sin }^2}x} \right)}}{{\left( {{{\sin }^4}x + {{\cos }^4}x} \right)}}\\& = - \cos 2x\\&\therefore \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx = \int { - \cos 2xdx = - \frac{{\sin 2x}}{2}} + C\end{align}\]

Chapter 7 Ex.7.ME Question 11

Integrate \(\frac{1}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}\)

 

Solution

 

\(\frac{1}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}\)

Multiplying and dividing by \(\sin \left( {a - b} \right)\), we get

\[\begin{align}\frac{1}{{\sin \left( {a - b} \right)}}\left[ {\frac{{\sin \left( {a - b} \right)}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}} \right]& = \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\frac{{\sin \left[ {\left( {x + a} \right) - \left( {x + b} \right)} \right]}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}} \right]\\& = \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\frac{{\sin \left( {x + a} \right)\cos \left( {x + b} \right) - \cos \left( {x + a} \right)\sin \left( {x + b} \right)}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}} \right]\\ &= \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\frac{{\sin \left( {x + a} \right)}}{{\cos \left( {x + a} \right)}} - \frac{{\sin \left( {x + b} \right)}}{{\cos \left( {x + b} \right)}}} \right]\\ &= \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\tan \left( {x + a} \right) - \tan \left( {x + b} \right)} \right]\\\int {\frac{1}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}} dx &= \frac{1}{{\sin \left( {a - b} \right)}}\int {\left[ {\tan \left( {x + a} \right) - \tan \left( {x + b} \right)} \right]} dx\\& = \frac{1}{{\sin \left( {a - b} \right)}}\left[ { - \log \left| {\cos \left( {x + a} \right)} \right| + \log \left| {\cos \left( {x + b} \right)} \right|} \right] + C\\ &= \frac{1}{{\sin \left( {a - b} \right)}}\log \left| {\frac{{\cos \left( {x + b} \right)}}{{\cos \left( {x + a} \right)}}} \right| + C\end{align}\]

Chapter 7 Ex.7.ME Question 12

Integrate \(\frac{{{x^3}}}{{\sqrt {1 - {x^8}} }}\)

 

Solution

 

\(\frac{{{x^3}}}{{\sqrt {1 - {x^8}} }}\)

Put, \({x^4} = t \Rightarrow \; 4{x^3}dx = dt\)

\[\begin{align} &\Rightarrow \; \int {\frac{{{x^3}}}{{\sqrt {1 - {x^8}} }}} dx = \frac{1}{4}\int {\frac{{dt}}{{\sqrt {1 - {t^2}} }}} \\ &= \frac{1}{4}{\sin ^{ - 1}}t + C\\ &= \frac{1}{4}{\sin ^{ - 1}}\left( {{x^4}} \right) + C\end{align}\]

Chapter 7 Ex.7.ME Question 13

Integrate \(\frac{{{e^x}}}{{\left( {1 + {e^x}} \right)\left( {2 + {e^x}} \right)}}\)

 

Solution

 

\(\frac{{{e^x}}}{{\left( {1 + {e^x}} \right)\left( {2 + {e^x}} \right)}}\)

Put \({e^x} = t \Rightarrow \; {e^x}dx = dt\)

\[\begin{align} &\Rightarrow \; \int {\frac{{{e^x}}}{{\left( {1 + {e^x}} \right)\left( {2 + {e^x}} \right)}}dx = \int {\frac{{dt}}{{\left( {t + 1} \right)\left( {t + 2} \right)}}} } \\ &= \int {\left[ {\frac{1}{{\left( {t + 1} \right)}} - \frac{1}{{\left( {t + 2} \right)}}} \right]dt} \\& = \log \left| {t + 1} \right| - \log \left| {t + 2} \right| + C\\ &= \log \left| {\frac{{t + 1}}{{t + 2}}} \right| + C\\ &= \log \left| {\frac{{1 + {e^x}}}{{2 + {e^x}}}} \right| + C\end{align}\]

Chapter 7 Ex.7.ME Question 14

Integrate \(\frac{1}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}\)

 

Solution

 

\[\begin{align}&\therefore \frac{1}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}} = \frac{{Ax + B}}{{\left( {{x^2} + 1} \right)}} + \frac{{Cx + D}}{{\left( {{x^2} + 4} \right)}}\\ &\Rightarrow \; 1 = \left( {Ax + B} \right)\left( {{x^2} + 4} \right) + \left( {Cx + D} \right)\left( {{x^2} + 1} \right)\\& \Rightarrow \; 1 = A{x^3} + 4Ax + B{x^2} + 4B + C{x^3} + Cx + D{x^2} + D\end{align}\]

Equating the coefficients of \({x^3},{x^2},x\)and constant term, we get

\[\begin{align}A + C &= 0\\B + D &= 0\\4A + C &= 0\\4B + D &= 1\end{align}\]

On solving these equations, we get

\[\begin{align}A &= 0\\B &= \frac{1}{3}\\C &= 0\\D &= - \frac{1}{3}\end{align}\]

From equation (\(1\)), we get

\[\begin{align}\frac{1}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}} &= \frac{1}{{3\left( {{x^2} + 1} \right)}} - \frac{1}{{3\left( {{x^2} + 4} \right)}}\\\int {\frac{1}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}} dx &= \frac{1}{3}\int {\frac{1}{{{x^2} + 1}}} dx - \frac{1}{3}\int {\frac{1}{{{x^2} + 4}}} dx\\ &= \frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{3}.\frac{1}{2}{\tan ^{ - 1}}\frac{x}{2} + C\\& = \frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\frac{x}{2} + C\end{align}\]

Chapter 7 Ex.7.ME Question 15

Integrate \({\cos ^3}x{e^{\log \sin x}}\)

 

Solution

 

\({\cos ^3}x{e^{\log \sin x}} = {\cos ^3}x \times \sin x\)

Let \(\cos x = t \Rightarrow \; - \sin xdx = dt\)

\[\begin{align}& \Rightarrow \; \int {{{\cos }^3}x{e^{\log \sin x}}} dx = \int {{{\cos }^3}x\sin xdx} \\ &= - \int {{t^3}dt} \\& = - \frac{{{t^4}}}{4} + C\\ &= - \frac{{{{\cos }^4}x}}{4} + C\end{align}\]

Chapter 7 Ex.7.ME Question 16

Integrate \({e^{3\log x}}{\left( {{x^4} + 1} \right)^{ - 1}}\)

 

Solution

 

\({e^{3\log x}}{\left( {{x^4} + 1} \right)^{ - 1}} = {e^{\log {x^3}}}{\left( {{x^4} + 1} \right)^{ - 1}} = \frac{{{x^3}}}{{\left( {{x^4} + 1} \right)}}\)

Let \({x^4} + 1 = t \Rightarrow \; 4{x^3}dx = dt\)

\[\begin{align}& \Rightarrow \; \int {{e^{3\log x}}{{\left( {{x^4} + 1} \right)}^{ - 1}}} dx = \int {\frac{{{x^3}}}{{\left( {{x^4} + 1} \right)}}} dx\\ &= \frac{1}{4}\int {\frac{{dt}}{t}} \\ &= \frac{1}{4}\log \left| t \right| + C\\& = \frac{1}{4}\log \left| {{x^4} + 1} \right| + C\\ &= \frac{1}{4}\log \left( {{x^4} + 1} \right) + C\end{align}\]

Chapter 7 Ex.7.ME Question 17

Integrate \(f'\left( {ax + b} \right){\left[ {f\left( {ax + b} \right)} \right]^n}\)

 

Solution

 

\(f'\left( {ax + b} \right){\left[ {f\left( {ax + b} \right)} \right]^n}\)

Put, \(f\left( {ax + b} \right) = t \Rightarrow \; af'\left( {ax + b} \right)dx = dt\)

\[\begin{align} &\Rightarrow \; \int {f'\left( {ax + b} \right){{\left[ {f\left( {ax + b} \right)} \right]}^n}} dx = \frac{1}{a}\int {{t^n}} dt\\ &= \frac{1}{a}\left[ {\frac{{{t^{n + 1}}}}{{n + 1}}} \right] = \frac{1}{{a\left( {n + 1} \right)}}{\left( {f\left( {ax + b} \right)} \right)^{n + 1}} + C\end{align}\]

Chapter 7 Ex.7.ME Question 18

Integrate \(\frac{1}{{\sqrt {{{\sin }^3}x\sin \left( {x + \alpha } \right)} }}\)

 

Solution

 

\[\begin{align}\frac{1}{{\sqrt {{{\sin }^3}x\sin \left( {x + \alpha } \right)} }} &= \frac{1}{{\sqrt {{{\sin }^3}x\left( {\sin x\cos \alpha + \cos x\sin \alpha } \right)} }}\\ &= \frac{1}{{\sqrt {{{\sin }^4}x\cos \alpha + {{\sin }^3}x\cos x\sin \alpha } }}\\ &= \frac{1}{{{{\sin }^2}x\sqrt {\cos \alpha + \cot x\sin \alpha } }} = \frac{{\cos e{c^2}x}}{{\sqrt {\cos \alpha + \cot x\sin \alpha } }}\end{align}\]

Put, \(\cos \alpha + \cot x\sin \alpha = t \Rightarrow \; - \cos e{c^2}x\sin \alpha dx = dt\)

\[\begin{align}\therefore \int {\frac{1}{{\sqrt {{{\sin }^3}x\sin \left( {x + \alpha } \right)} }}} dx &= \int {\frac{{\cos e{c^2}x}}{{\sqrt {\cos \alpha + \cot x\sin \alpha } }}} dx\\ &= \frac{{ - 1}}{{\sin \alpha }}\int {\frac{{dt}}{{\sqrt t }}} \\& = \frac{{ - 1}}{{\sin \alpha }}\left[ {2\sqrt t } \right] + C\\ &= \frac{{ - 1}}{{\sin \alpha }}\left[ {2\sqrt {\cos \alpha + \cot x\sin \alpha } } \right] + C\\ &= \frac{{ - 2}}{{\sin \alpha }}\sqrt {\cos \alpha + \frac{{\cos x\sin \alpha }}{{\sin x}}} + C\\ &= \frac{{ - 2}}{{\sin \alpha }}\sqrt {\frac{{\sin x\cos \alpha + \cos x\sin \alpha }}{{\sin x}}} + C = \frac{{ - 2}}{{\sin \alpha }}\sqrt {\frac{{\sin \left( {x + \alpha } \right)}}{{\sin x}}} + C\end{align}\]

Chapter 7 Ex.7.ME Question 19

Integrate \(\frac{{{{\sin }^{ - 1}}\sqrt x - {{\cos }^{ - 1}}\sqrt x }}{{{{\sin }^{ - 1}}\sqrt x + {{\cos }^{ - 1}}\sqrt x }},x \in \left[ {0,1} \right]\)

 

Solution

 

Let \(I = \frac{{{{\sin }^{ - 1}}\sqrt x - {{\cos }^{ - 1}}\sqrt x }}{{{{\sin }^{ - 1}}\sqrt x + {{\cos }^{ - 1}}\sqrt x }}dx\)

As we know that, \({\sin ^{ - 1}}\sqrt x + {\cos ^{ - 1}}\sqrt x = \frac{\pi }{2}\)

\[\begin{align} &\Rightarrow \; I = \int {\frac{{\left( {\frac{\pi }{2} - {{\cos }^{ - 1}}\sqrt x } \right) - {{\cos }^{ - 1}}\sqrt x }}{{\frac{\pi }{2}}}} dx\\ &= \frac{2}{\pi }\int {\left( {\frac{\pi }{2} - 2{{\cos }^{ - 1}}\sqrt x } \right)} dx\\ &= \frac{2}{\pi }.\frac{\pi }{2}\int {1.dx} - \frac{4}{\pi }\int {{{\cos }^{ - 1}}} \sqrt x dx\\& = x - \frac{4}{\pi }\int {{{\cos }^{ - 1}}} \sqrt x dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\end{align}\]

Let \({I_1} = \int {{{\cos }^{ - 1}}\sqrt x } dx\)

Also, let \(\sqrt x = t \Rightarrow \; dx = 2tdt\)

\[\begin{align} \Rightarrow \; {I_1} &= 2\int {{{\cos }^{ - 1}}} t.tdt\\& = 2\left[ {{{\cos }^{ - 1}}t.\frac{{{t^2}}}{2} - \int {\frac{{ - 1}}{{\sqrt {1 - {t^2}} }}.\frac{{{t^2}}}{2}} dt} \right]\\ &= {t^2}{\cos ^{ - 1}}t + \int {\frac{{{t^2}}}{{\sqrt {1 - {t^2}} }}} dt\\ &= {t^2}{\cos ^{ - 1}}t - \int {\frac{{1 - {t^2} - 1}}{{\sqrt {1 - {t^2}} }}} dt\\& = {t^2}{\cos ^{ - 1}}t - \int {\sqrt {1 - {t^2}} dt + \int {\frac{1}{{\sqrt {1 - {t^2}} }}} } dt\\ &= {t^2}{\cos ^{ - 1}}t - \frac{1}{2}\sqrt {1 - {t^2}} - \frac{1}{2}{\sin ^{ - 1}}t + {\sin ^{ - 1}}t\\ &= {t^2}{\cos ^{ - 1}}t - \frac{1}{2}\sqrt {1 - {t^2}} + \frac{1}{2}{\sin ^{ - 1}}t\end{align}\]

From equation (1), we get

\[\begin{align}I &= x - \frac{4}{\pi }\left[ {{t^2}{{\cos }^{ - 1}}t - \frac{t}{2}\sqrt {1 - {t^2}} + \frac{1}{2}{{\sin }^{ - 1}}t} \right]\\ &= x - \frac{4}{\pi }\left[ {x{{\cos }^{ - 1}}\sqrt x - \frac{{\sqrt x }}{2}\sqrt {1 - x} + \frac{1}{2}{{\sin }^{ - 1}}\sqrt x } \right]\\&x - \frac{4}{\pi }\left[ {x\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}\sqrt x } \right) - \frac{{\sqrt {x - {x^2}} }}{2} + \frac{\pi }{2}{{\sin }^{ - 1}}\sqrt x } \right]\\ &= x - 2x + \frac{{4x}}{\pi }{\sin ^{ - 1}}\sqrt x + \frac{2}{\pi }\sqrt {x - {x^2}} - \frac{2}{\pi }{\sin ^{ - 1}}\sqrt x\\&- x + \frac{2}{\pi }\left[ {\left( {2x - 1} \right){{\sin }^{ - 1}}\sqrt x } \right] + \frac{2}{\pi }\sqrt {x - {x^2}} + C\\ &= \frac{{2\left( {2x - 1} \right)}}{\pi }{\sin ^ - }\sqrt x + \frac{2}{\pi }\sqrt {x - {x^2}} - x + C\end{align}\]

Chapter 7 Ex.7.ME Question 20

Integrate \(\sqrt {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \)

 

Solution

 

\(I = \sqrt {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} \)

Put, \(x = {\cos ^2}\theta \Rightarrow \; dx = - 2\sin \theta \cos \theta d\theta \)

\[\begin{align}I& = \sqrt {\frac{{1 - \cos \theta }}{{1 + \cos \theta }}} \left( { - 2\sin \theta \cos \theta } \right)d\theta = - \int {\sqrt {\frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2{{\cos }^2}\frac{\theta }{2}}}} } \sin 2\theta d\theta \\ &= - 2\int {\frac{{\sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2}}}} \left( {2\sin \frac{\theta }{2}\cos \frac{\theta }{2}} \right)\cos \theta d\theta \\ &= - 4\int {{{\sin }^2}\frac{\theta }{2}} \cos \theta d\theta \\& = - 4\int {{{\sin }^2}\frac{\theta }{2}} \left( {2{{\cos }^2}\frac{\theta }{2} - 1} \right)d\theta \\ &= - 4\int {\left( {2{{\sin }^2}\frac{\theta }{2}{{\cos }^2}\frac{\theta }{2} - {{\sin }^2}\frac{\theta }{2}} \right)} d\theta \\ &= - 8\int {{{\sin }^2}\frac{\theta }{2}.{{\cos }^2}\frac{\theta }{2}d\theta } + 4\int {{{\sin }^2}\frac{\theta }{2}} d\theta \\ &= - 2\int {{{\sin }^2}\frac{\theta }{2}} d\theta + 4\int {{{\sin }^2}\frac{\theta }{2}} d\theta \\ &= - 2\int {\left( {\frac{{1 - \cos 2\theta }}{2}} \right)} d\theta + 4\int {\frac{{1 - \cos \theta }}{2}} d\theta \\ &= - 2\left[ {\frac{\theta }{2} - \frac{{\sin 2\theta }}{4}} \right] + 4\left[ {\frac{\theta }{2} - \frac{{\sin \theta }}{2}} \right] + C\\ &= - \theta + \frac{{\sin 2\theta }}{2} + 2\theta - 2\sin \theta + C\\ &= \theta + \frac{{\sin 2\theta }}{2} + 2\sin \theta + C\\ &= \theta + \frac{{2\sin \theta \cos \theta }}{2} - 2\sin \theta + C\\ &= \theta + \sqrt {1 - {{\cos }^2}\theta } .\cos \theta - 2\sqrt {1 - {{\cos }^2}\theta } + C\\ &= {\cos ^{ - 1}}\sqrt x + \sqrt {1 - x} .\sqrt x - 2\sqrt {1 - x} + C\\& = - 2\sqrt {1 - x} + {\cos ^{ - 1}}\sqrt x + \sqrt {x\left( {1 - x} \right)} + C\\ &= - 2\sqrt {1 - x} + {\cos ^{ - 1}}\sqrt x + \sqrt {x - {x^2}} + C\end{align}\]

Chapter 7 Ex.7.ME Question 21

Integrate \(\frac{{2 + \sin 2x}}{{1 + \cos 2x}}{e^x}\)

 

Solution

 

\[\begin{align}I &= \int {\left( {\frac{{2 + \sin 2x}}{{1 + \cos 2x}}} \right)} {e^x}\\& = \int {\left( {\frac{{2 + 2\sin x\cos x}}{{2{{\cos }^2}x}}} \right)} {e^x}\\ &= \int {\left( {\frac{{1 + \sin x\cos x}}{{{{\cos }^2}x}}} \right)} {e^x}\\ &= \int {\left( {{{\sec }^2}x + \tan x} \right)} {e^x}\end{align}\]

Let \(f\left( x \right) = \tan x \Rightarrow \; f'\left( x \right) = {\sec ^2}x\)

\[\begin{align}\therefore I &= \int {\left( {f\left( x \right) + f'\left( x \right)} \right){e^x}} dx\\ &= {e^x}f\left( x \right) + C\\& = {e^x}\tan x + C\end{align}\]

Chapter 7 Ex.7.ME Question 22

Integrate \(\frac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}\)

 

Solution

 

Let \(\frac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}} = \frac{A}{{\left( {x + 1} \right)}} + \frac{B}{{{{\left( {x + 1} \right)}^2}}} + \frac{C}{{\left( {x + 2} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\, \cdots \left( 1 \right)\)

\[\begin{align} &\Rightarrow \; {x^2} + x + 1 = A\left( {x + 1} \right)\left( {x + 2} \right) + B\left( {x + 2} \right) + C\left( {{x^2} + 2x + 1} \right)\\ &\Rightarrow \; {x^2} + x + 1 = A\left( {{x^2} + 3x + 2} \right) + B\left( {x + 2} \right) + C\left( {{x^2} + 2x + 1} \right)\\& \Rightarrow \; {x^2} + x + 1 = \left( {A + C} \right){x^2} + \left( {3A + B + 2C} \right)x + \left( {2A + 2B + C} \right)\end{align}\]

Equating the coefficients of \({x^2},x\)and constant term, we get

\[\begin{align}A + C &= 1\\3A + B + 2C &= 1\\2A + 2B + C& = 1\end{align}\]

On solving these equations, we get

\[\begin{align}A &= - 2\\B& = 1\\C &= 3\end{align}\]

From equation (\(1\)), we get

\[\begin{align}\frac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}& = \frac{{ - 2}}{{\left( {x + 1} \right)}} + \frac{3}{{\left( {x + 2} \right)}} + \frac{1}{{{{\left( {x + 1} \right)}^2}}}\\\int {\frac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}} dx &= - 2\int {\frac{1}{{x + 1}}} dx + 3\int {\frac{1}{{\left( {x + 2} \right)}}dx + \int {\frac{1}{{{{\left( {x + 1} \right)}^2}}}} } dx\\ &= - 2\log \left| {x + 1} \right| + 3\log \left| {x + 2} \right| - \frac{1}{{\left( {x + 1} \right)}} + C\end{align}\]

Chapter 7 Ex.7.ME Question 23

Integrate \({\tan ^{ - 1}}\sqrt {\frac{{1 - x}}{{1 + x}}} \)

 

Solution

 

\(I = {\tan ^{ - 1}}\sqrt {\frac{{1 - x}}{{1 + x}}} dx\)

Let \(x = \cos \theta \Rightarrow \; dx = - \sin \theta d\theta \)

\[\begin{align}I& = \int {{{\tan }^{ - 1}}\sqrt {\frac{{1 - \cos \theta }}{{1 + \cos \theta }}} } \left( { - \sin \theta } \right)d\theta \\ &= - \int {{{\tan }^{ - 1}}\sqrt {\frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2{{\cos }^2}\frac{\theta }{2}}}\sin \theta } } d\theta = - \int {{{\tan }^{ - 1}}\tan \frac{\theta }{2}} \sin \theta d\theta \\ &= - \frac{1}{2}\int {\theta .\sin \theta d\theta } = - \frac{1}{2}\left[ {\theta .\left( { - \cos \theta } \right) - \int {1.\left( { - \cos \theta } \right)d\theta } } \right]\\&= - \frac{1}{2}\left[ { - \theta \cos \theta + \sin \theta } \right]\\ &= \frac{1}{2}\theta \cos \theta - \frac{1}{2}\sin \theta \\ &= \frac{1}{2}{\cos ^{ - 1}}x.x - \frac{1}{2}\sqrt {1 - {x^2}} + C = \frac{x}{2}{\cos ^{ - 1}}x - \frac{1}{2}\sqrt {1 - {x^2}} + C\\ &= \frac{1}{2}\left( {x{{\cos }^{ - 1}}x - \sqrt {1 - {x^2}} } \right) + C\end{align}\]

Chapter 7 Ex.7.ME Question 24

Integrate \(\frac{{\sqrt {{x^2} + 1} \left[ {\log \left( {{x^2} + 1} \right) - 2\log x} \right]}}{{{x^4}}}\)

 

Solution

 

\[\begin{align}\frac{{\sqrt {{x^2} + 1} \left[ {\log \left( {{x^2} + 1} \right) - 2\log x} \right]}}{{{x^4}}} &= \frac{{\sqrt {{x^2} + 1} }}{{{x^4}}}\left[ {\log \left( {{x^2} + 1} \right) - \log {x^2}} \right]\\ &= \frac{{\sqrt {{x^2} + 1} }}{{{x^4}}}\left[ {\log \left( {\frac{{{x^2} + 1}}{{{x^2}}}} \right)} \right]\\& = \frac{{\sqrt {{x^2} + 1} }}{{{x^4}}}\log \left( {1 + \frac{1}{{{x^2}}}} \right)\\& = \frac{1}{{{x^3}}}\sqrt {\frac{{{x^2} + 1}}{{{x^2}}}} \log \left( {1 + \frac{1}{{{x^2}}}} \right)\\& = \frac{1}{{{x^3}}}\sqrt {1 + \frac{1}{{{x^2}}}} \log \left( {1 + \frac{1}{{{x^2}}}} \right)\end{align}\]

Let \(1 + \frac{1}{{{x^2}}} = t \Rightarrow \; \frac{{ - 2}}{{{x^3}}}dx = dt\)

\[\begin{align}\therefore I &= \int {\frac{1}{{{x^3}}}\sqrt {1 + \frac{1}{{{x^2}}}} } \log \left( {1 + \frac{1}{{{x^2}}}} \right)dx\\ &= - \frac{1}{2}\int {\sqrt t } \log tdt = - \frac{1}{2}\int {{t^{\frac{1}{2}}}\log tdt}\end{align}\]

Using integration by parts, we get

\[\begin{align}I &= - \frac{1}{2}\left[ {\log t.\int {{t^{\frac{1}{2}}}} dt - \left\{ {\left( {\frac{d}{{dt}}\log t} \right)\int {{t^{\frac{1}{2}}}dt} } \right\}dt} \right]\\ &= - \frac{1}{2}\left[ {\log t.\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}} - \int {\frac{1}{t}.\frac{{{t^{\frac{3}{2}}}}}{{\frac{3}{2}}}dt} } \right]\\ &= - \frac{1}{2}\left[ {\frac{2}{3}{t^{\frac{3}{2}}}\log t - \frac{2}{3}\int {{t^{\frac{1}{2}}}dt} } \right]\\ &= - \frac{1}{2}\left[ {\frac{2}{3}{t^{\frac{3}{2}}}\log t - \frac{4}{9}{t^{\frac{3}{2}}}} \right]\\ &= - \frac{1}{3}{t^{\frac{3}{2}}}\log t + \frac{2}{9}{t^{\frac{3}{2}}}\\ &= - \frac{1}{3}{t^{\frac{3}{2}}}\left[ {\log t - \frac{2}{3}} \right]\\ &= - \frac{1}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{\frac{3}{2}}}\left[ {\log \left( {1 + \frac{1}{{{x^2}}}} \right) - \frac{2}{3}} \right] + C\end{align}\]

Chapter 7 Ex.7.ME Question 25

Evaluate the definite integral \(\int_{\frac{\pi }{2}}^\pi {{e^x}} \left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx\)

 

Solution

 

\[\begin{align}I& = \int_{\frac{\pi }{2}}^\pi {{e^x}} \left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx\\ &= \int_{\frac{\pi }{2}}^\pi {{e^x}} \left( {\frac{{1 - 2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\sin }^2}\frac{x}{2}}}} \right)dx = \int_{\frac{\pi }{2}}^\pi {{e^x}\left( {\frac{{\cos e{c^2}\frac{x}{2}}}{2} - \cot \frac{x}{2}} \right)} dx\end{align}\]

Let \(f\left( x \right) = - \cot \frac{x}{2}\)

\[\begin{align} \Rightarrow \; f'\left( x \right)& = - \left( {\frac{1}{2}\cos e{c^2}\frac{x}{2}} \right) = \frac{1}{2}\cos e{c^2}\frac{x}{2}\\\therefore I &= \int_{\frac{\pi }{2}}^\pi {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)} dx\\ &= \left[ {{e^x}f\left( x \right)dx} \right]_{\frac{\pi }{2}}^\pi \\ &= - \left[ {{e^x}\cot \frac{x}{2}} \right]_{\frac{\pi }{2}}^\pi \\ &= - \left[ {{e^x} \times \cot \frac{\pi }{2} - {e^{\frac{\pi }{2}}} \times \cot \frac{\pi }{4}} \right]\\ &= - \left[ {{e^\pi } \times 0 - {e^{\frac{\pi }{2}}} \times 1} \right]\\& = {e^{\frac{\pi }{2}}}\end{align}\]

Chapter 7 Ex.7.ME Question 26

Evaluate the definite integral \(\int_0^{\frac{\pi }{4}} {\frac{{\sin x\cos x}}{{{{\cos }^4}x + {{\sin }^4}x}}dx} \)

 

Solution

 

Let \(I = \int_0^{\frac{\pi }{4}} {\frac{{\sin x\cos x}}{{{{\cos }^4}x + {{\sin }^4}x}}dx} \)

\[\begin{align}& \Rightarrow \; I = \int_0^{\frac{\pi }{4}} {\frac{{\frac{{\left( {\sin x\cos x} \right)}}{{{{\cos }^4}x}}}}{{\frac{{\left( {{{\cos }^4}x + {{\sin }^4}x} \right)}}{{{{\cos }^4}x}}}}dx} \\& \Rightarrow \; I = \int_0^{\frac{\pi }{4}} {\frac{{\tan x{{\sec }^2}x}}{{1 + {{\tan }^4}x}}} dx\end{align}\]

Put, \({\tan ^2}x = t \Rightarrow \; 2\tan x{\sec ^2}xdx = dt\)

When \(x = 0,t = 0\)and when \(x = \frac{\pi }{4},t = 1\)

\[\begin{align}\therefore I &= \frac{1}{2}\int_0^1 {\frac{{dt}}{{1 + {t^2}}}} = \frac{1}{2}\left[ {{{\tan }^{ - 1}}t} \right]_0^1\\ &= \frac{1}{2}\left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}0} \right]\\ &= \frac{1}{2}\left[ {\frac{\pi }{4}} \right]\\& = \frac{\pi }{8}\end{align}\]

Chapter 7 Ex.7.ME Question 27

Evaluate the definite integral \(\int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^2}x}}{{{{\cos }^2}x + 4{{\sin }^2}x}}dx} \)

 

Solution

 

Consider, \(I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^2}x}}{{{{\cos }^2}x + 4{{\sin }^2}x}}dx} \)

\[\begin{align} &\Rightarrow \; I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^2}x}}{{{{\cos }^2}x + 4\left( {1 - {{\cos }^2}x} \right)}}dx} \\& \Rightarrow \; I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^2}x}}{{{{\cos }^2}x + 4 - 4{{\cos }^2}x}}dx} \\ &\Rightarrow \; I = \frac{{ - 1}}{3}\int_0^{\frac{\pi }{2}} {\frac{{ - 3{{\cos }^2}x}}{{4 - 3{{\cos }^2}x}}\;} dx\\& \Rightarrow \; I = \frac{{ - 1}}{3}\int_0^{\frac{\pi }{2}} {\frac{{4 - 3{{\cos }^2}x - 4}}{{4 - 3{{\cos }^2}x}}\;} dx\\& \Rightarrow \; I = \frac{{ - 1}}{3}\int_0^{\frac{\pi }{2}} {\frac{{4 - 3{{\cos }^2}x}}{{4 - 3{{\cos }^2}x}}} \;dx + \frac{1}{3}\int_0^{\frac{\pi }{2}} {\frac{4}{{4 - 3{{\cos }^2}x}}} \;dx\\& \Rightarrow \; I = \frac{{ - 1}}{3}\int_0^{\frac{\pi }{2}} {1dx} + \frac{1}{3}\int_0^{\frac{\pi }{2}} {\frac{{4{{\sec }^2}x}}{{4{{\sec }^2}x - 3}}} dx\\ &\Rightarrow \; I = - \frac{1}{3}\left[ x \right]_0^{\frac{\pi }{2}} + \frac{1}{3}\int_0^{\frac{\pi }{2}} {\frac{{4{{\sec }^2}x}}{{4\left( {1 + {{\tan }^2}x} \right) - 3}}} dx\\& \Rightarrow \; I = - \frac{\pi }{6} + \frac{2}{3}\int_0^{\frac{\pi }{2}} {\frac{{2{{\sec }^2}x}}{{1 + 4{{\tan }^2}x}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\end{align}\]

Consider, \(\int_0^{\frac{\pi }{2}} {\frac{{2{{\sec }^2}x}}{{1 + 4{{\tan }^2}x}}} dx\)

Put, \(2\tan x = t \Rightarrow \; 2{\sec ^2}xdx = dt\)

When \(x = 0,t = 0\)and when \(x = \frac{\pi }{2},t = \infty \)

\[\begin{align} \Rightarrow \; \int_0^{\frac{\pi }{2}} {\frac{{2{{\sec }^2}x}}{{1 + 4{{\tan }^2}x}}} dx& = \int_0^\infty {\frac{{dt}}{{1 + {t^2}}}} \\ &= \left[ {{{\tan }^{ - 1}}t} \right]_0^\infty \\ &= \left[ {{{\tan }^{ - 1}}\left( \infty \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right]\\& = \frac{\pi }{2}\end{align}\]

Therefore, from (\(1\)), we get

\(I = - \frac{\pi }{6} + \frac{2}{3}\left[ {\frac{\pi }{2}} \right] = \frac{\pi }{3} - \frac{\pi }{6} = \frac{\pi }{6}\)

Chapter 7 Ex.7.ME Question 28

Evaluate the definite integral \(\int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{\sin x + \cos x}}{{\sqrt {\sin 2x} }}} dx\)

 

Solution

 

Consider, \(I = \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{\sin x + \cos x}}{{\sqrt {\sin 2x} }}} dx\)

\[\begin{align}&\Rightarrow \; I = \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{\sin x + \cos x}}{{\sqrt { - \left( { - \sin 2x} \right)} }}} dx \Rightarrow \; I = \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{\sin x + \cos x}}{{\sqrt { - \left( { - 1 + 1 - 2\sin \cos x} \right)} }}} dx\\& \Rightarrow \; I = \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{\sin x + \cos x}}{{\sqrt {1 - {{\left( {\sin x - \cos x} \right)}^2}} }}} dx\end{align}\]

Let \(\left( {\sin x - \cos x} \right) = t \Rightarrow \; \left( {\sin x + \cos x} \right)dx = dt\)

When \(x = \frac{\pi }{6},t = \left( {\frac{{1 - \sqrt 3 }}{2}} \right)\) and when \(x = \frac{\pi }{3},t = \left( {\frac{{\sqrt 3 - 1}}{2}} \right)\)

\[\begin{align}&I = \int_{\frac{{1 - \sqrt 3 }}{2}}^{\frac{{\sqrt 3 - 1}}{2}} {\frac{{dt}}{{\sqrt {1 - {t^2}} }}} \\ &\Rightarrow \; I = \int_{ - \left( {\frac{{1 + \sqrt 3 }}{2}} \right)}^{\frac{{\sqrt 3 - 1}}{2}} {\frac{{dt}}{{\sqrt {1 - {t^2}} }}}\end{align}\]

As \(\frac{1}{{\sqrt {1 - {{\left( { - t} \right)}^2}} }} = \frac{1}{{\sqrt {1 - {t^2}} }}\), therefore, \(\frac{1}{{\sqrt {1 - {t^2}} }}\)is an even function

\(\int_{ - a}^a {f\left( x \right)dx} = 2\int_0^a {f\left( x \right)dx} \)

We know that if \(f\left( x \right)\) is an even function, then

\[\begin{align} \Rightarrow \; I& = 2\int_0^{\frac{{\sqrt 3 - 1}}{0}} {\frac{{dt}}{{\sqrt {1 - {t^2}} }}} \\& = \left[ {2{{\sin }^{ - 1}}t} \right]_0^{\frac{{\sqrt 3 - 1}}{2}}\\& = 2{\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - 1}}{2}} \right)\end{align}\]

Chapter 7 Ex.7.ME Question 29

Evaluate the definite integral \(\int_0^1 {\frac{{dx}}{{\sqrt {1 + x} - \sqrt x }}} \)

 

Solution

 

Consider, \(I = \int_0^1 {\frac{{dx}}{{\sqrt {1 + x} - \sqrt x }}} \)

\[\begin{align}I &= \int_0^1 {\frac{1}{{\left( {\sqrt {1 + x} - \sqrt x } \right)}} \times \frac{{\left( {\sqrt {1 + x} + \sqrt x } \right)}}{{\left( {\sqrt {1 + x} + \sqrt x } \right)}}} dx \hfill \\&= \int_0^1 {\frac{{\left( {\sqrt {1 + x} + \sqrt x } \right)}}{{1 + x - x}}} dx \hfill \\&= \int_0^1 {\sqrt {1 + x} dx} + \int_0^1 {\sqrt x dx} \hfill \\&= \left[ {\frac{2}{3}{{\left( {1 + x} \right)}^{\frac{3}{2}}}} \right]_0^1 + \left[ {\frac{2}{3}{{\left( x \right)}^{\frac{3}{2}}}} \right]_0^1 \hfill \\&= \frac{2}{3}\left[ {{{\left( 2 \right)}^{\frac{3}{2}}} - 1} \right] + \frac{2}{3}\left[ 1 \right] \hfill \\&= \frac{2}{3}{\left( 2 \right)^{\frac{3}{2}}} = \frac{{2.2\sqrt 2 }}{3} \hfill \\&= \frac{{4\sqrt 2 }}{3} \hfill \\\end{align} \]

Chapter 7 Ex.7.ME Question 30

Evaluate the definite integral \(\int_0^{\frac{\pi }{4}} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}} dx\)

 

Solution

 

Consider, \(I = \int_0^{\frac{\pi }{4}} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}} dx\)

Put, \(\sin x - \cos x = t \Rightarrow \; \left( {\cos x + \sin x} \right)dx = dt\)

When \(x = 0,t = - 1\) and when \(x = \frac{\pi }{4},t = 0\)

\[\begin{align} &\Rightarrow \; {\left( {\sin x - \cos x} \right)^2} = {t^2}\\& \Rightarrow \; {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x = {t^2}\\ &\Rightarrow \; 1 - \sin 2x = {t^2}\\ &\Rightarrow \; \sin 2x = 1 - {t^2}\\\therefore I &= \int_{ - 1}^0 {\frac{{dt}}{{9 + 16\left( {1 - {t^2}} \right)}}} \\ &= \int_{ - 1}^0 {\frac{{dt}}{{9 + 16 - 16{t^2}}}} \\ &= \int_{ - 1}^0 {\frac{{dt}}{{25 - 16{t^2}}}} = \int_{ - 1}^0 {\frac{{dt}}{{{{\left( 5 \right)}^2} - {{\left( {4t} \right)}^2}}}} \\ &= \frac{1}{4}\left[ {\frac{1}{{2\left( 5 \right)}}\log \left| {\frac{{5 + 4t}}{{5 - 4t}}} \right|} \right]_{ - 1}^0\\ &= \frac{1}{{40}}\left[ {\log \left( 1 \right) - \log \left| {\frac{1}{9}} \right|} \right]\\& = \frac{1}{{40}}\log 9\end{align}\]

Chapter 7 Ex.7.ME Question 31

Evaluate the definite integral \(\int_0^{\frac{\pi }{2}} {\sin 2x{{\tan }^{ - 1}}\left( {\sin x} \right)} dx\)

 

Solution

 

Consider, \(I = \int_0^{\frac{\pi }{2}} {\sin 2x{{\tan }^{ - 1}}\left( {\sin x} \right)} dx = \int_0^{\frac{\pi }{2}} {2\sin x\cos x{{\tan }^{ - 1}}\left( {\sin x} \right)} dx\)

Put, \(\sin x = t \Rightarrow \; \cos xdx = dt\)

When \(x = 0,t = 0\)and when \(\)

\( \Rightarrow \; I = 2\int_0^1 {t{{\tan }^{ - 1}}\left( t \right)dt} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\)

Consider \(\int {t.{{\tan }^{ - 1}}tdt = {{\tan }^{ - 1}}t\int {tdt - \int {\left\{ {\frac{d}{{dt}}\left( {{{\tan }^{ - 1}}t} \right)\int {tdt} } \right\}} } } dt\)

\[\begin{align} &= {\tan ^{ - 1}}t.\frac{{{t^2}}}{2} - \int {\frac{1}{{1 + {t^2}}}.\frac{{{t^2}}}{2}} dt\\ &= \frac{{{t^2}{{\tan }^{ - 1}}t}}{2} - \frac{1}{2}\int {\frac{{{t^2} + 1 - 1}}{{1 + {t^2}}}} dt\\ &= \frac{{{t^2}{{\tan }^{ - 1}}t}}{2} - \frac{1}{2}\int {1.dt} + \frac{1}{2}\int {\frac{1}{{1 + {t^2}}}dt} \\ &= \frac{{{t^2}{{\tan }^{ - 1}}t}}{2} - \frac{1}{2}t + \frac{1}{2}{\tan ^{ - 1}}t\end{align}\]

\[\begin{align}{\text{From equation }}\left( {\text{1}} \right),{\text{ we get}}\\ &\Rightarrow \; 2\int_0^1 {t.{{\tan }^{ - 1}}tdt} = 2\left[ {\frac{{{t^2}{{\tan }^{ - 1}}t}}{2} - \frac{t}{2} + \frac{1}{2}{{\tan }^{ - 1}}t} \right]_0^1\\ &= \left[ {\frac{\pi }{4} - 1 + \frac{\pi }{4}} \right] = \frac{\pi }{2} - 1\end{align}\]

Chapter 7 Ex.7.ME Question 32

Evaluate the definite integral \(\int_0^\pi {\frac{{x\tan x}}{{\sec x + \tan x}}} dx\)

 

Solution

 

Let \(\int_0^\pi {\frac{{x\tan x}}{{\sec x + \tan x}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\)

\[\begin{align}I &= \int_0^\pi {\left\{ {\frac{{\left( {\pi - x} \right)\tan \left( {\pi - x} \right)}}{{\sec \left( {\pi - x} \right) + \tan \left( {\pi - x} \right)}}} \right\}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} } \right)\\& \Rightarrow \; I = \int_0^\pi {\left\{ {\frac{{ - \left( {\pi - x} \right)\tan x}}{{ - \left( {\sec x + \tan x} \right)}}} \right\}dx} \\ &\Rightarrow \; I = \int_0^\pi {\frac{{\left( {\pi - x} \right)\tan x}}{{\left( {\sec x + \tan x} \right)}}dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\end{align}\]

Adding (1) and (2), we get

\[\begin{align}2I &= \int_0^\pi {\frac{{\pi \tan x}}{{\sec x + \tan x}}} dx \Rightarrow \; 2I = \pi \int_0^\pi {\frac{{\frac{{\sin x}}{{\cos x}}}}{{\frac{1}{{\cos x}} + \frac{{\sin x}}{{\cos x}}}}} dx\\& \Rightarrow \; 2I = \pi \int_0^\pi {\frac{{\sin x + 1 - 1}}{{1 + \sin x}}} dx\\ &\Rightarrow \; 2I = \pi \int_0^\pi {1.dx} - \pi \int_0^\pi {\frac{1}{{1 + \sin x}}} dx\\& \Rightarrow \; 2I = \pi \int_0^\pi {1.dx} - \pi \int_0^\pi {\frac{{\left( {1 - \sin x} \right)}}{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}} dx\\ &\Rightarrow \; 2I = \pi \left[ x \right]_0^\pi - \pi \int_0^\pi {\frac{{1 - \sin x}}{{{{\cos }^2}x}}dx} \\& \Rightarrow \; 2I = {\pi ^2} - \pi \int_0^\pi {\left( {{{\sec }^2}x - \tan x\sec x} \right)dx} \\ &\Rightarrow \; 2I = {\pi ^2} - \pi \left[ {\tan x - \sec x} \right]_0^\pi \\& \Rightarrow \; 2I = {\pi ^2} - \pi \left[ {\tan \pi - \sec \pi - \tan 0 + \sec 0} \right]\\ &\Rightarrow \; 2I = {\pi ^2} - \pi \left[ {0 - \left( { - 1} \right) - 0 + 1} \right]\\ &\Rightarrow \; 2I = {\pi ^2} - 2\pi \\& \Rightarrow \; 2I = \pi \left( {\pi - 2} \right)\\I &= \frac{\pi }{2}\left( {\pi - 2} \right)\end{align}\]

Chapter 7 Ex.7.ME Question 33

Evaluate the definite integral \(\int_1^4 {\left[ {\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right|} \right]} dx\)

 

Solution

 

Consider, \(I = \int_1^4 {\left[ {\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right|} \right]} dx\)

\( \Rightarrow \; I = \int_1^4 {\left| {x - 1} \right|dx} + \int_1^4 {\left| {x + 2} \right|} dx + \int_1^4 {\left| {x + 3} \right|} dx\)

\(I = {I_1} + {I_2} + {I_3}\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\)

Where, \({I_1} = \int_1^4 {\left| {x - 1} \right|dx} ,{I_2} = \int_1^4 {\left| {x + 2} \right|} dx{\text{ and }}{I_3} = \int_1^4 {\left| {x + 3} \right|} dx\)

\({I_1} = \int_1^4 {\left| {x - 1} \right|dx} \)

\[\begin{align}&\left( {x - 1} \right) \ge 0\,\,for\,\,1 \le x \le 4\\&\therefore {I_1} = \int_1^4 {\left( {x - 1} \right)dx} \\& \Rightarrow \; {I_1} = \left[ {\frac{{{x^2}}}{2} - x} \right]_1^4\\& \Rightarrow \; {I_1} = \left[ {8 - 4 - \frac{1}{2} + 1} \right] = \frac{9}{2}\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\end{align}\]

\({I_2} = \int_1^4 {\left| {x - 2} \right|} dx\)

\[\begin{align}&x - 2 \ge 0\,\,for\,\,2 \le x \le 4\,\,{\text{and }}x - 2 \le 0\,\,for\,\,1 \le x \le 2\\&\therefore {I_2} = \int_1^2 {\left( {2 - x} \right)dx + \int_2^4 {\left( {x - 2} \right)dx} } \\ &\Rightarrow \; {I_2} = \left[ {2x - \frac{{{x^2}}}{2}} \right]_1^2 + \left[ {\frac{{{x^2}}}{2} - 2x} \right]_2^4 \Rightarrow \; {I_2} = \left[ {4 - 2 - 2 + \frac{1}{2}} \right] + \left[ {8 - 8 - 2 + 4} \right]\\ &\Rightarrow \; {I_2} = \frac{1}{2} + 2 = \frac{5}{2}\\& \Rightarrow \; {I_3} = \int_1^4 {\left| {x - 3} \right|dx}\end{align}\]

\[\begin{align}&x - 3 \ge 0\,\,for\,\,3 \le x \le 4\,\,{\text{and}}\,\,x - 3 \le 0\,\,for\,\,1 \le x \le 2\\&\therefore {I_3} = \int_1^3 {\left( {3 - x} \right)} dx + \int_3^4 {\left( {x - 3} \right)dx} \\ &\Rightarrow \; {I_3} = \left[ {3 - \frac{{{x^2}}}{2}} \right]_1^3 + \left[ {\frac{{{x^2}}}{2} - 3x} \right]_3^4\\& \Rightarrow \; {I_3} = \left[ {9 - \frac{9}{2} - 3 + \frac{1}{2}} \right] + \left[ {8 - 12 - \frac{9}{2} + 9} \right]\\ &\Rightarrow \; {I_3} = \left[ {6 - 4} \right] + \left[ {\frac{1}{2}} \right] = \frac{5}{2}\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 4 \right)\end{align}\]

From equations (\(1\)), (\(2\)), (\(3\)) and (\(4\)), we get

\(I = \frac{9}{2} + \frac{5}{2} + \frac{5}{2} = \frac{{19}}{2}\)

Chapter 7 Ex.7.ME Question 34

Prove: \(\int_1^3 {\frac{{dx}}{{{x^2}\left( {x + 1} \right)}} = \frac{2}{3} + \log \frac{2}{3}} \)

 

Solution

 

Consider, \(\int_1^3 {\frac{{dx}}{{{x^2}\left( {x + 1} \right)}}} \)

Let, \(\frac{1}{{{x^2}\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}}\)

\[\begin{align} &\Rightarrow \; 1 = Ax\left( {x + 1} \right) + B\left( {x + 1} \right) + C\left( {{x^2}} \right)\\ &\Rightarrow \; 1 = A{x^2} + Ax + Bx + B + C{x^2}\end{align}\]

Equating the coefficients of \({x^2},x\)and constant terms, we get

\[\begin{align}A + C &= 0\\A + B &= 0\\B &= 1\end{align}\]

On solving these equations, we get

\[\begin{align}A &= - 1\\C &= 1\\B &= 1\end{align}\]

\[\begin{align}\therefore \frac{1}{{{x^2}\left( {x + 1} \right)}}& = \frac{{ - 1}}{x} + \frac{1}{{{x^2}}} + \frac{1}{{\left( {x + 1} \right)}}\\ &\Rightarrow \; I = \int_1^3 {\left\{ { - \frac{1}{x} + \frac{1}{{{x^2}}} + \frac{1}{{\left( {x + 1} \right)}}} \right\}} dx = \left[ { - \log x - \frac{1}{x} + \log \left( {x + 1} \right)} \right]_1^3\\& = \left[ {\log \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{x}} \right]_1^3 = \log \left( {\frac{4}{3}} \right) - \frac{1}{3} - \log \left( {\frac{2}{1}} \right) + 1\\& = \log 4 - \log 3 - \log 2 + \frac{2}{3}\\ &= \log 2 - \log 3 + \frac{2}{3}\\& = \log \left( {\frac{2}{3}} \right) + \frac{2}{3}\end{align}\]

Hence proved.

Chapter 7 Ex.7.ME Question 35

Prove: \(\int_0^1 {x{e^x}dx = 1} \)

 

Solution

 

Let \(I = \int_0^1 {x{e^x}dx} \)

Using integration by parts, we get

\[\begin{align}I &= x\int_0^1 {{e^x}dx} - \int_0^1 {\left\{ {\left( {\frac{d}{{dx}}\left( x \right)} \right)\int {{e^x}dx} } \right\}dx} \\& = \left[ {x{e^x}} \right]_0^1 - \int_0^1 {{e^x}dx} \\ &= \left[ {x{e^x}} \right]_0^1 - \left[ {{e^x}} \right]_0^1\\ &= e - e + 1\\ &= 1\end{align}\]

Hence proved.

Chapter 7 Ex.7.ME Question 36

Prove: \(\int_{ - 1}^1 {{x^{17}}{{\cos }^4}xdx} = 0\)

 

Solution

 

Consider, \(I = \int_{ - 1}^1 {{x^{17}}{{\cos }^4}xdx} \)

Let \(f\left( x \right) = {x^{17}}{\cos ^4}x\)

\( \Rightarrow \; f\left( { - x} \right) = {\left( { - x} \right)^{17}}{\cos ^4}\left( { - x} \right) = - {x^{17}}{\cos ^4}x = - f\left( x \right)\)

\(f\left( x \right)\)is an odd function.

We know that if \(f\left( x \right)\) is an odd function, then \(\int_{ - a}^a {f\left( x \right)dx = 0} \)

\(\therefore I = \int_{ - 1}^1 {{x^{17}}{{\cos }^4}xdx = 0} \)

Hence proved.

Chapter 7 Ex.7.ME Question 37

Prove: \(\int_0^{\frac{\pi }{2}} {{{\sin }^3}xdx} = \frac{2}{3}\)

 

Solution

 

Consider, \(I = \int_0^{\frac{\pi }{2}} {{{\sin }^3}xdx} \)

\[\begin{align}I &= \int_0^{\frac{\pi }{2}} {{{\sin }^2}x.\sin xdx} \\& = \int_0^{\frac{\pi }{2}} {\left( {1 - {{\cos }^2}x} \right)\sin xdx} \\& = \int_0^{\frac{\pi }{2}} {\sin xdx} - \int_0^{\frac{\pi }{2}} {{{\cos }^2}x.\sin xdx} \\ &= \left[ { - \cos x} \right]_0^{\frac{\pi }{2}} + \left[ {\frac{{{{\cos }^3}x}}{3}} \right]_0^{\frac{\pi }{2}}\\ &= 1 + \frac{1}{3}\left[ { - 1} \right] = 1 - \frac{1}{3} = \frac{2}{3}\end{align}\]

Hence proved.

Chapter 7 Ex.7.ME Question 38

Prove: \(\int_0^{\frac{\pi }{4}} {2{{\tan }^3}xdx} = 1 - \log 2\)

 

Solution

 

Consider, \(I = \int_0^{\frac{\pi }{4}} {2{{\tan }^3}xdx} \)

\[\begin{align}I& = \int_0^{\frac{\pi }{4}} {2{{\tan }^2}x.\tan xdx} = 2\int_0^{\frac{\pi }{4}} {\left( {{{\sec }^2} - 1} \right)\tan xdx} \\ &= 2\int_0^{\frac{\pi }{4}} {{{\sec }^2}x\tan xdx - 2\int_0^{\frac{\pi }{4}} {\tan xdx} } \\ &= 2\left[ {\frac{{{{\tan }^2}x}}{2}} \right]_0^{\frac{\pi }{4}} + 2\left[ {\log \cos x} \right]_0^{\frac{\pi }{4}} = 1 + 2\left[ {\log \cos \frac{\pi }{4} - \log \cos 0} \right]\\ &= 1 + 2\left[ {\log \frac{1}{{\sqrt 2 }} - \log 1} \right] = 1 - \log 2 - \log 1 = 1 - \log 2\end{align}\]

Hence proved.

Chapter 7 Ex.7.ME Question 39

Prove: \(\int_0^1 {{{\sin }^{ - 1}}x} dx = \frac{\pi }{2} - 1\)

 

Solution

 

Let \(\int_0^1 {{{\sin }^{ - 1}}x} dx\)

\( \Rightarrow \; I = \int_0^1 {{{\sin }^{ - 1}}x} .1.dx\)

Using integration by parts, we get

\[\begin{align}I &= \left[ {{{\sin }^{ - 1}}x.x} \right]_0^1 - \int_0^1 {\frac{1}{{\sqrt {1 - {x^2}} }}} xdx\\ &= \left[ {x{{\sin }^{ - 1}}x} \right]_0^1 + \frac{1}{2}\int_0^1 {\frac{{\left( { - 2x} \right)}}{{\sqrt {1 - {x^2}} }}dx}\end{align}\]

Put, \(1 - {x^2} = t \Rightarrow \; - 2xdx = dt\)

When \(x = 0,t = 1\)and when \(x = 1,t = 0\)

\[\begin{align}I &= \left[ {x{{\sin }^{ - 1}}x} \right]_0^1 + \frac{1}{2}\int_0^1 {\frac{{dt}}{{\sqrt t }}} \\ &= \left[ {x{{\sin }^{ - 1}}x} \right]_0^1 + \frac{1}{2}\left[ {2\sqrt t } \right]_1^0\\& = {\sin ^{ - 1}}\left( 1 \right) + \left[ { - \sqrt 1 } \right]\\ &= \frac{\pi }{2} - 1\end{align}\]

Hence proved.

Chapter 7 Ex.7.ME Question 40

Evaluate \(\int_0^1 {{e^{2 - 3x}}dx} \) as a limit of a sum.

 

Solution

 

Let \(I = \int_0^1 {{e^{2 - 3x}}dx} \)

We know that,

\(\int_a^b {f\left( x \right)dx = \left( {b - a} \right)\mathop {\lim }\limits_{n \to \infty } } \frac{1}{n}\left[ {f\left( a \right) + f\left( {a + h} \right) + \ldots + f\left( {a + \left( {n - 1} \right)h} \right)} \right]\)

Where, \(h = \frac{{b - a}}{n}\)

Here, \(a = 0,\;b = 1\)and \(f\left( x \right) = {e^{2 - 3x}}\)

\[\begin{align} \Rightarrow \; h &= \frac{{1 - 0}}{n} = \frac{1}{n}\\\therefore \int_0^1 {{e^{2 - 3x}}} dx& = \left( {1 - 0} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {f\left( 0 \right) + f\left( {0 + h} \right) + \ldots + f\left( {0 + \left( {n - 1} \right)h} \right)} \right]\\& = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{e^2} + {e^{2 - 3x}} + \ldots + {e^{2 - 3\left( {n - 1} \right)h}}} \right] = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{e^2}\left\{ {1 + {e^{ - 3h}} + {e^{ - 6h}} + {e^{ - 9h}} + \ldots + {e^{ - 3\left( {n - 1} \right)h}}} \right\}} \right]\\ &= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{e^2}\left\{ {\frac{{1 - {{\left( {{e^{ - 3h}}} \right)}^n}}}{{1 - \left( {{e^{ - 3h}}} \right)}}} \right\}} \right] = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {{e^2}\left\{ {\frac{{1 - {e^{ - \frac{3}{n}n}}}}{{1 - {e^{ - \frac{3}{n}}}}}} \right\}} \right]\\& = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\frac{{{e^2}\left( {1 - {e^{ - 3}}} \right)}}{{1 - {e^{ - \frac{3}{n}}}}}} \right] = {e^2}\left( {{e^{ - 3}} - 1} \right)\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\frac{1}{{{e^{ - \frac{3}{n}}} - 1}}} \right]\\ &= {e^2}\left( {{e^{ - 3}} - 1} \right)\mathop {\lim }\limits_{n \to \infty } \left( { - \frac{1}{3}} \right)\left[ {\frac{{ - \frac{3}{n}}}{{{e^{ - \frac{3}{n}}} - 1}}} \right] = \frac{{{e^2}\left( {{e^{ - 3}} - 1} \right)}}{3}\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\frac{{ - 3}}{n}}}{{{e^{ - \frac{3}{n}}} - 1}}} \right]\\ &= \frac{{ - {e^2}\left( {{e^{ - 3}} - 1} \right)}}{3}\left( 1 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\\ &= \frac{{ - {e^{ - 1}} + {e^2}}}{3}\\ &= \frac{1}{3}\left( {{e^2} - \frac{1}{e}} \right)\end{align}\]

Chapter 7 Ex.7.ME Question 41

\(\int {\frac{{dx}}{{{e^x} + {e^{ - x}}}}} \) is equal to

\(\begin{align}A.\,\,{\tan ^{ - 1}}\left( {{e^x}} \right) + C\\B.\,\,{\tan ^{ - 1}}\left( {{e^{ - x}}} \right) + C\\C.\,\,\log \left( {{e^x} - {e^{ - x}}} \right) + C\\D.\,\,\log \left( {{e^x} + {e^{ - x}}} \right) + C\end{align}\)

 

Solution

 

Consider, \(I = \int {\frac{{dx}}{{{e^x} + {e^{ - x}}}}dx} = \int {\frac{{{e^x}}}{{{e^{2x}} + 1}}} dx\)

Put, \({e^x} = t \Rightarrow \; {e^x}dx = dt\)

\(\therefore I = \int {\frac{{dt}}{{1 + {t^2}}}} \)

\[\begin{align} &= {\tan ^{ - 1}}t + C\\& = {\tan ^{ - 1}}\left( {{e^x}} \right) + C\end{align}\]

Thus, the correct option is A.

Chapter 7 Ex.7.ME Question 42

\(\int {\frac{{\cos 2x}}{{{{\left( {\sin x + \cos x} \right)}^2}}}dx} \) is equal to

\(\begin{align}&A.\,\,\frac{{ - 1}}{{\sin x + \cos x}} + C\\&B.\,\,\log \left| {\sin x + \cos x} \right| + C\\&C.\,\,\log \left| {\sin x - \cos x} \right| + C\\&D.\,\,\frac{1}{{{{\left( {\sin x + \cos x} \right)}^2}}} + C\end{align}\)

 

Solution

 

Consider, \(I = \int {\frac{{\cos 2x}}{{{{\left( {\sin x + \cos x} \right)}^2}}}dx} \, \Rightarrow \; I = \int {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{\left( {\sin x + \cos x} \right)}}dx} \)

\( = \int {\frac{{\left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right)}}{{{{\left( {\sin x + \cos x} \right)}^2}}}} dx = \int {\frac{{\cos x - \sin x}}{{\cos x + \sin x}}} dx\)

Let \(\cos x + \sin x = t \Rightarrow \; \left( {\cos x - \sin x} \right)dx = dt\)

\[\begin{align}\therefore I& = \int {\frac{{dt}}{t}} \\ &= \log \left| t \right| + C\\ &= \log \left| {\cos x + \sin x} \right| + C\end{align}\]

Thus, the correct option is B.

Chapter 7 Ex.7.ME Question 43

If \(f\left( {a + b - x} \right) = f\left( x \right)\), then \(\int_a^b {xf\left( x \right)dx} \) is equal to

\(\begin{align}&A.\,\,\frac{{a + b}}{2}\int_a^b {f\left( {b - x} \right)} dx\\&B.\,\,\frac{{a + b}}{2}\int_a^b {f\left( {b + x} \right)} dx\\&C.\,\,\frac{{b - a}}{2}\int_a^b {f\left( x \right)} dx\\&D.\,\,\frac{{a + b}}{2}\int_a^b {f\left( x \right)} dx\end{align}\)

 

Solution

 

Consider, \(I = \int_a^b {xf\left( x \right)dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\)

\[\begin{align}I &= \int_a^b {\left( {a + b - x} \right)f\left( {a + b - x} \right)} dx \quad \left( {\int_a^b {f\left( x \right)} dx = \int_a^b {f\left( {a + b - x} \right)} dx} \right)\\& \Rightarrow \; I = \int_a^b {\left( {a + b - x} \right)f\left( x \right)dx} \\& \Rightarrow \; I = \left( {a + b} \right)\int_a^b {f\left( x \right)dx - I} ......\;\left( {{\text{Using equation }}\left( {} \right)} \right)\\ &\Rightarrow \; I + I = \left( {a + b} \right)\int_a^b {f\left( x \right)dx} \\ &\Rightarrow \; 2I = \left( {a + b} \right)\int_a^b {f\left( x \right)dx} \\ &\Rightarrow \; I = \left( {\frac{{a + b}}{2}} \right)\int_a^b {f\left( x \right)dx}\end{align}\]

Thus, the correct option is D.

Chapter 7 Ex.7.ME Question 44

The value of \(\int_0^1 {{{\tan }^{ - 1}}\left( {\frac{{2x - 1}}{{1 + x - {x^2}}}} \right)dx} \) is

A. \(1\)

B. \(0\)

C. \(-1\)

D. \(\frac{\pi }{4}\)

 

Solution

 

Consider, \(I = \int_0^1 {{{\tan }^{ - 1}}\left( {\frac{{2x - 1}}{{1 + x - {x^2}}}} \right)dx} \)

\[\begin{align} &\Rightarrow \; I = \int_0^1 {{{\tan }^{ - 1}}\left( {\frac{{x - \left( {1 - x} \right)}}{{1 + x\left( {1 - x} \right)}}} \right)} dx\\ &\Rightarrow \; I = \int_0^1 {\left[ {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}\left( {1 - x} \right)} \right]dx\,\,\,\,\, \ldots \left( 1 \right)} \\& \Rightarrow \; I = \int_0^1 {\left[ {{{\tan }^{ - 1}}\left( {1 - x} \right) - {{\tan }^{ - 1}}\left( {1 - 1 + x} \right)} \right]} dx\\ &\Rightarrow \; I = \int_0^1 {\left[ {{{\tan }^{ - 1}}\left( {1 - x} \right) - {{\tan }^{ - 1}}x} \right]} dx\\ &\Rightarrow \; I = \int_0^1 {\left[ {{{\tan }^{ - 1}}\left( {1 - x} \right) - {{\tan }^{ - 1}}\left( x \right)} \right]} dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\end{align}\]

Adding (1) and (2), we get

\[\begin{align} &\Rightarrow \; 2I = \int_0^1 {\left( {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}\left( {1 - x} \right) - {{\tan }^{ - 1}}\left( {1 - x} \right) - {{\tan }^{ - 1}}x} \right)} dx\\ & \Rightarrow \; 2I = 0\\ &\Rightarrow \; I = 0\end{align}\]

Thus, the correct option is B.

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