NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.9

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Chapter 7 Ex.7.9 Question 1

Evaluate the definite integral \(\int_{ - 1}^1 {\left( {x + 1} \right)dx} \)

Solution

Let \(I = \int_{ - 1}^1 {\left( {x + 1} \right)dx} \)

\(\int {\left( {x + 1} \right)dx} = \frac{{{x^2}}}{2} + x = F\left( x \right)\)

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 1 \right) - F\left( { - 1} \right)\\ &= \left( {\frac{1}{2} + 1} \right) - \left( {\frac{1}{2} - 1} \right)\\ &= \frac{1}{2} + 1 - \frac{1}{2} + 1\\ &= 2\end{align}\]

Chapter 7 Ex.7.9 Question 2

Evaluate the definite integral \(\int_2^3 {\frac{1}{x}dx} \)

Solution

Let \(I = \int_2^3 {\frac{1}{x}dx} \)

\(\int {\frac{1}{x}dx} = \log \left| x \right| = F\left( x \right)\)

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 3 \right) - F\left( 2 \right)\\ &= \log \left| 3 \right| - \log \left| 2 \right| = \log \frac{3}{2}\end{align}\]

Chapter 7 Ex.7.9 Question 3

Evaluate the definite integral \(\int_1^2 {\left( {4{x^3} - 5{x^2} + 6x + 9} \right)dx} \)

Solution

Let \(I = \int_1^2 {\left( {4{x^3} - 5{x^2} + 6x + 9} \right)dx} \)

\[\begin{align}\int {\left( {4{x^3} - 5{x^2} + 6x + 9} \right)dx} &= {4\left( {\frac{{{x^4}}}{4}} \right) - 5\left( {\frac{{{x^3}}}{3}} \right) + 6\left( {\frac{{{x^2}}}{2}} \right)} + 9\left( x \right)\\ &= {x^4} - \frac{{5{x^3}}}{3} + 3{x^2} + 9x = F\left( x \right)\end{align}\]

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 2 \right) - F\left( 1 \right)\\I& = \left\{ {{2^4} - \frac{{5{{\left( 2 \right)}^3}}}{3} + 3{{\left( 2 \right)}^2} + 9\left( 2 \right)} \right\} - \left\{ {{{\left( 1 \right)}^4} - \frac{{5{{\left( 1 \right)}^3}}}{3} + 3{{\left( 1 \right)}^2} + 9\left( 1 \right)} \right\}\\ &= \left( {16 - \frac{{40}}{3} + 12 + 18} \right) - \left( {1 - \frac{5}{3} + 3 + 9} \right)\\ &= 16 - \frac{{40}}{3} + 12 + 18 - 1 + \frac{5}{3} - 3 - 9\\ &= 33 - \frac{{35}}{3}\\ &= \frac{{99 - 35}}{3}\\& = \frac{{64}}{3}\end{align}\]

Chapter 7 Ex.7.9 Question 4

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

Solution

\[\begin{align}{\rm{Let}}\;I &= \int_0^{\frac{\pi }{4}} {\sin 2x\;dx} \\\int {\sin 2x\;dx} &= \left( {\frac{{ - \cos 2x}}{2}} \right) = F\left( x \right)\end{align}\]

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( {\frac{\pi }{4}} \right) - F\left( 0 \right)\\& = - \frac{1}{2}\left[ {\cos 2\left( {\frac{\pi }{4}} \right) - \cos 0} \right] = - \frac{1}{2}\left[ {\cos \left( {\frac{\pi }{2}} \right) - \cos 0} \right]\\&= - \frac{1}{2}\left[ {0 - 1} \right]\\ &= \frac{1}{2}\end{align}\]

Chapter 7 Ex.7.9 Question 5

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

Solution

Let \(I = \int_0^{\frac{\pi }{2}} {\cos 2x\;dx} \)

\(\int {\cos 2x\;dx = \left( {\frac{{\sin 2x}}{2}} \right)} = F\left( x \right)\)

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( {\frac{\pi }{2}} \right) - F\left( 0 \right)\\& = \frac{1}{2}\left[ {\sin 2\left( {\frac{\pi }{2}} \right) - \sin 0} \right] = \frac{1}{2}\left[ {\sin \pi - \sin 0} \right]\\& = \frac{1}{2}\left[ {0 - 0} \right] = 0\end{align}\]

Chapter 7 Ex.7.9 Question 6

Evaluate the definite integral \(\int_4^5 {{e^x}dx} \)

Solution

Let \(I = \int_4^5 {{e^x}dx} \)

\(\int {{e^x}} dx = {e^x} = F\left( x \right)\)

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 5 \right) - F\left( 4 \right)\\& = {e^5} - {e^4}\\& = {e^4}\left( {e - 1} \right)\end{align}\]

Chapter 7 Ex.7.9 Question 7

Evaluate the definite integral \(\int_0^{\frac{\pi }{4}} {\tan x\;dx} \)

Solution

Let \(I = \int_0^{\frac{\pi }{4}} {\tan x\;dx} \)

\(\int {\tan x\;dx = - \log \left| {\cos x} \right|} = F\left( x \right)\)

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( {\frac{\pi }{4}} \right) - F\left( 0 \right)\\ &= - \log \left| {\cos \frac{\pi }{4}} \right| + \log \left| {\cos 0} \right| = - \log \left| {\frac{1}{{\sqrt 2 }}} \right| + \log \left| 1 \right|\\ &= - \log {\left( 2 \right)^{ - \frac{1}{2}}}\\ &= \frac{1}{2}\log 2\end{align}\]

Chapter 7 Ex.7.9 Question 8

Evaluate the definite integral \(\int_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\cos ec\;x\;dx} \)

Solution

Let \(I = \int_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\cos ec\;x\;dx} \)

\(\int {\cos ec\;x\;dx} = \log \left| {\cos ec\;x - \cot x} \right| = F\left( x \right)\)

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( {\frac{\pi }{4}} \right) - F\left( {\frac{\pi }{6}} \right)\\& = \log \left| {\cos ec\frac{\pi }{4} - \cot \frac{\pi }{4}} \right| - \log \left| {\cos ec\frac{\pi }{6} - \cot \frac{\pi }{6}} \right|\\\log \left| {\sqrt 2 - 1} \right| - \log \left| {2 - \sqrt 3 } \right| &= \log \left( {\frac{{\sqrt 2 - 1}}{{2 - \sqrt 3 }}} \right)\end{align}\]

Chapter 7 Ex.7.9 Question 9

Evaluate the definite integral \(\int_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \)

Solution

Let \(I = \int_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \)

\(\int {\frac{{dx}}{{\sqrt {1 - {x^2}} }} = {{\sin }^{ - 1}}x = F\left( x \right)} \)

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 1 \right) - F\left( 0 \right)\\ &= {\sin ^{ - 1}}\left( 1 \right) - {\sin ^{ - 1}}\left( 0 \right)\\ &= \frac{\pi }{2} - 0\\& = \frac{\pi }{2}\end{align}\]

Chapter 7 Ex.7.9 Question 10

Evaluate the definite integral \(\int_0^1 {\frac{{dx}}{{1 + {x^2}}}} \)

Solution

Let \(I = \int_0^1 {\frac{{dx}}{{1 + {x^2}}}} \)

\(\int {\frac{{dx}}{{1 + {x^2}}} = {{\tan }^{ - 1}}x} = F\left( x \right)\)

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 1 \right) - F\left( 0 \right)\\ &= {\tan ^{ - 1}}\left( 1 \right) - {\tan ^{ - 1}}\left( 0 \right)\\ &= \frac{\pi }{4}\end{align}\]

Chapter 7 Ex.7.9 Question 11

Evaluate the definite integral \(\int_2^3 {\frac{{dx}}{{{x^2} - 1}}} \)

Solution

\[\begin{align}{\rm{Let }}I& = \int_2^3 {\frac{{dx}}{{{x^2} - 1}}} \\\int {\frac{{dx}}{{{x^2} - 1}}} &= \frac{1}{2}\log \left| {\frac{{x - 1}}{{x + 1}}} \right| = F\left( x \right)\end{align}\]

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 3 \right) - F\left( 2 \right)\\& = \frac{1}{2}\left[ {\log \left| {\frac{{3 - 1}}{{3 + 1}}} \right| - \log \left| {\frac{{2 - 1}}{{2 + 1}}} \right|} \right] = \frac{1}{2}\left[ {\log \left| {\frac{2}{4}} \right| - \log \left| {\frac{1}{3}} \right|} \right]\\ &= \frac{1}{2}\left[ {\log \frac{1}{2} - \log \frac{1}{3}} \right]\\& = \frac{1}{2}\left[ {\log \frac{3}{2}} \right]\end{align}\]

Chapter 7 Ex.7.9 Question 12

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

Solution

Let \(I = \int_0^{\frac{\pi }{2}} {{{\cos }^2}x\;dx} \)

\(\int {{{\cos }^2}x\;dx} = \int {\left( {\frac{{1 + \cos 2x}}{2}} \right)} dx = \frac{x}{2} + \frac{{\sin 2x}}{4} = \frac{1}{2}\left( {x + \frac{{\sin 2x}}{2}} \right) = F\left( x \right)\)

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= \left[ {F\left( {\frac{\pi }{2}} \right) - F\left( 0 \right)} \right] = \frac{1}{2}\left[ {\left( {\frac{\pi }{2} + \frac{{\sin \pi }}{2}} \right) - \left( {0 + \frac{{\sin \pi }}{2}} \right)} \right]\\ &= \frac{1}{2}\left[ {\frac{\pi }{2} + 0 - 0 - 0} \right]\\ &= \frac{\pi }{4}\end{align}\]

Chapter 7 Ex.7.9 Question 13

Evaluate the definite integral \(\int_2^3 {\frac{x}{{{x^2} + 1}}} dx\)

Solution

Let \(I = \int_2^3 {\frac{x}{{{x^2} + 1}}} dx\)

\(\int {\frac{x}{{{x^2} + 1}}} dx = \frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}} dx = \frac{1}{2}\log \left( {1 + {x^2}} \right) = F\left( x \right)\)

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 3 \right) - F\left( 2 \right)\\ &= \frac{1}{2}\left[ {\log \left( {1 + {{\left( 3 \right)}^2}} \right) - \log \left( {1 + {{\left( 2 \right)}^2}} \right)} \right]\\ &= \frac{1}{2}\left[ {\log \left( {10} \right) - \log \left( 5 \right)} \right]\\ &= \frac{1}{2}\log \left( {\frac{{10}}{5}} \right) = \frac{1}{2}\log 2\end{align}\]

Chapter 7 Ex.7.9 Question 14

Evaluate the definite integral \(\int_0^1 {\frac{{2x + 3}}{{5{x^2} + 1}}} dx\)

Solution

Let \(I = \int_0^1 {\frac{{2x + 3}}{{5{x^2} + 1}}} dx\)

\[\begin{align}\int {\frac{{2x + 3}}{{5{x^2} + 1}}} dx &= \frac{1}{5}\int {\frac{{5\left( {2x + 3} \right)}}{{5{x^2} + 1}}} dx = \frac{1}{5}\int {\frac{{10x + 15}}{{5{x^2} + 1}}} dx\\[10pt] &= \frac{1}{5}\int {\frac{{10x}}{{5{x^2} + 1}}} dx + 3\int {\frac{1}{{5{x^2} + 1}}} dx\\[10pt] &= \frac{1}{5}\int {\frac{{10x}}{{5{x^2} + 1}}} dx + 3\int {\frac{1}{{5\left( {{x^2} + \frac{1}{5}} \right)}}} dx \\[10pt] &= \frac{1}{5}\log \left( {5{x^2} + 1} \right) + \frac{3}{5}.\frac{1}{{\frac{1}{{\sqrt 5 }}}}{\tan ^{ - 1}}\frac{x}{{\frac{1}{{\sqrt 5 }}}}\\[5pt] &= \frac{1}{5}\log \left( {5{x^2} + 1} \right) + \frac{3}{{\sqrt 5 }}{\tan ^{ - 1}}\left( {\sqrt 5 } \right)x\\[5pt] &= F\left( x \right)\end{align}\]

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 1 \right) - F\left( 0 \right)\\ &= \left\{ {\frac{1}{5}\log \left( {5 + 1} \right) + \frac{3}{{\sqrt 5 }}{{\tan }^{ - 1}}\left( {\sqrt 5 } \right)} \right\} - \left\{ {\frac{1}{5}\log \left( {5 \times 0 + 1} \right) + \frac{3}{{\sqrt 5 }}{{\tan }^{ - 1}}\left( 0 \right)} \right\}\\ &= \frac{1}{5}\log 6 + \frac{3}{{\sqrt 5 }}{\tan ^{ - 1}}\sqrt 5\end{align}\]

Chapter 7 Ex.7.9 Question 15

Evaluate the definite integral \(\int_0^1 {x{e^{{x^2}}}} dx\)

Solution

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

Put, \({x^2} = t \Rightarrow 2x\;dx = dt\)

As \(x \to 0,\;t \to 0\) and as\;\(x \to 1,\;t \to 1\)

\[\begin{align}\therefore I &= \frac{1}{2}\int_0^1 {{e^t}dt} \\\frac{1}{2}\int {{e^t}dt}& = \frac{1}{2}{e^t} = F\left( t \right)\end{align}\]

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 1 \right) - F\left( 0 \right)\\ &= \frac{1}{2}e - \frac{1}{2}{e^0}\\ &= \frac{1}{2}\left( {e - 1} \right)\end{align}\]

Chapter 7 Ex.7.9 Question 16

Evaluate the definite integral \(\int_1^2 {\frac{{5{x^2}}}{{{x^2} + 4x + 3}}} dx\)

Solution

Let \(I = \int_1^2 {\frac{{5{x^2}}}{{{x^2} + 4x + 3}}} dx\)

Dividing \(5{x^2}{\rm\;{ by }}\;{x^2} + 4x + 3\), we get

\[\begin{align}I &= \int_1^2 {\left\{ {5 - \frac{{20x + 15}}{{{x^2} + 4x + 3}}} \right\}} dx\\ &= \int_1^2 {5dx} - \int_1^2 {\frac{{20x + 15}}{{{x^2} + 4x + 3}}} dx\\ &= \left[ {5x} \right]_1^2 - \int_1^2 {\frac{{20x + 15}}{{{x^2} + 4x + 3}}} dx\\I = 5 - {I_1},\,\,{\rm{where }}\;I &= \int_1^2 {\frac{{20x + 15}}{{{x^2} + 4x + 3}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\end{align}\]

Let \(\begin{align} 20x + 15 &= A \frac{d}{{dx}}\left( {{x^2} + 4x + 3} \right) + B \\&= 2Ax + \left( {4A + B} \right) \end{align}\)

Equating the coefficients of x and constant term, we get

\(A = 10\;{\rm{and }}\;B = - 25\)

Let \({x^2} + 4x + 3 = t\)

\[\begin{align} &\Rightarrow \left( {2x + 4} \right)dx = dt\\ &\Rightarrow {I_1} = 10\int {\frac{{dt}}{t}} - 25\int {\frac{{dx}}{{{{\left( {x + 2} \right)}^2} - {1^2}}}} \\& = 10\log t - 25\left[ {\frac{1}{2}\log \left( {\frac{{x + 2 - 1}}{{x + 2 + 1}}} \right)} \right] = \left[ {10\log \left( {{x^2} + 4x + 3} \right)} \right]_1^2 - 25\left[ {\frac{1}{2}\log \left( {\frac{{x + 1}}{{x + 3}}} \right)} \right]_1^2\\ &= \left[ {10\log 15 - 10\log 8} \right] - 25\left[ {\frac{1}{2}\log \frac{3}{5} - \frac{1}{2}\log \frac{2}{4}} \right]\\ &= \left[ {10\log \left( {5 \times 3} \right) - 10\log \left( {4 \times 2} \right)} \right] - \frac{{25}}{2}\left[ {\log 3 - \log 5 - \log 2 + \log 4} \right]\\ &= \left[ {10\log 5 + 10\log 3 - 10\log 4 - 10\log 2} \right] - \frac{{25}}{2}\left[ {\log 3 - \log 5 - \log 2 + \log 4} \right]\\ &= \left[ {10 + \frac{{25}}{2}} \right]\log 5 + \left[ { - 10 - \frac{{25}}{2}} \right]\log 4 + \left[ {10 - \frac{{25}}{2}} \right]\log 3 + \left[ { - 10 + \frac{{25}}{2}} \right]\log 2\\ &= \frac{{45}}{2}\log 5 - \frac{{45}}{2}\log 4 - \frac{5}{2}\log 3 + \frac{5}{2}\log 2\\& = \frac{{45}}{2}\log \frac{5}{4} - \frac{5}{2}\log \frac{3}{2}\end{align}\]

Substituting the value \({I_1}\;\)n (1), we get

\[\begin{align}I &= 5 - \left[ {\frac{{45}}{2}\log \frac{5}{4} - \frac{5}{2}\log \frac{3}{2}} \right]\\ &= 5 - \frac{5}{2}\left[ {9\log \frac{5}{4} - \log \frac{3}{2}} \right]\end{align}\]

Chapter 7 Ex.7.9 Question 17

Evaluate the definite integral \(\int_0^{\frac{\pi }{4}} {\left( {2{{\;\sec }^2}x + {x^3} + 2} \right)dx} \)

Solution

Let \(I = \int_0^{\frac{\pi }{4}} {\left( {2{{\;\sec }^2}x + {x^3} + 2} \right)dx} \)

\(\int {\left( {2{{\;\sec }^2}x + {x^3} + 2} \right)} dx = 2\tan x + \frac{{{x^4}}}{4} + 2x = F\left( x \right)\)

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( {\frac{\pi }{4}} \right) - F\left( 0 \right) = \left\{ {\left( {2\tan \frac{\pi }{4} + \frac{1}{4}{{\left( {\frac{\pi }{4}} \right)}^4} + 2\left( {\frac{\pi }{4}} \right)} \right) - \left( {2\tan 0 + 0 + 0} \right)} \right\}\\& = 2\tan \frac{\pi }{4} + \frac{{{\pi ^4}}}{{{4^5}}} + \frac{\pi }{2}\\ &= 2 + \frac{\pi }{2} + \frac{{{\pi ^4}}}{{1024}}\end{align}\]

Chapter 7 Ex.7.9 Question 18

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

Solution

Let \(I = \int_0^\pi {\left( {{{\sin }^2}\frac{x}{2} - {{\cos }^2}\frac{x}{2}} \right)} dx = - \int_0^\pi {\left( {{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}} \right)} dx\)

\[\begin{align}&= - \int_0^\pi {\cos x\;dx} \hfill \\&\int {\cos x\;dx = \sin x = F\left( x \right)} \hfill \\\end{align} \]

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( \pi \right) - F\left( 0 \right)\\& = \sin \pi - \sin 0\\ &= 0\end{align}\]

Chapter 7 Ex.7.9 Question 19

Evaluate the definite integral \(\int_0^2 {\frac{{6x + 3}}{{{x^2} + 4}}} dx\)

Solution

Let \(I = \int_0^2 {\frac{{6x + 3}}{{{x^2} + 4}}} dx\)

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

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 2 \right) - F\left( 0 \right)\\ &= \left\{ {3\log \left( {{2^2} + 4} \right) + \frac{3}{2}{{\tan }^{ - 1}}\left( {\frac{2}{2}} \right)} \right\} - \left\{ {3\log \left( {0 + 4} \right) + \frac{3}{2}{{\tan }^{ - 1}}\left( {\frac{0}{2}} \right)} \right\}\\ &= 3\log 8 + \frac{3}{2}{\tan ^{ - 1}}1 - 3\log 4 - \frac{3}{2}{\tan ^{ - 1}}0\\ &= 3\log 8 + \frac{3}{2}\left( {\frac{\pi }{4}} \right) - 3\log 4 - 0\\ &= 3\log \left( {\frac{8}{4}} \right) + \frac{{3\pi }}{8}\\ &= 3\log 2 + \frac{{3\pi }}{8}\end{align}\]

Chapter 7 Ex.7.9 Question 20

Evaluate the definite integral \(\int_0^1 {\left( {x{e^x} + \sin \frac{{\pi x}}{4}} \right)dx} \)

Solution

Let \(I = \int_0^1 {\left( {x{e^x} + \sin \frac{{\pi x}}{4}} \right)dx} \)

\[\begin{align}\int_0^1 {\left( {x{e^x} + \sin \frac{{\pi x}}{4}} \right)dx} &= x\int {{e^x}dx} - \int {\left\{ {\left( {\frac{d}{{dx}}x} \right)\int {{e^x}dx} } \right\}dx + \left\{ {\frac{{ - \cos \frac{{\pi x}}{4}}}{{\frac{\pi }{4}}}} \right\}} \\ &= x{e^x} - \int {{e^x}dx} - \frac{4}{\pi }\cos \frac{{\pi x}}{4}\\ &= x{e^x} - {e^x} - \frac{4}{\pi }\cos \frac{{\pi x}}{4}\\ &= F\left( x \right)\end{align}\]

Using second fundamental theorem of calculus, we get

\[\begin{align}I &= F\left( 1 \right) - F\left( 0 \right)\\ &= \left( {1.{e^1} - {e^1} - \frac{4}{\pi }\cos \frac{\pi }{4}} \right) - \left( {0.{e^0} - {e^0} - \frac{4}{\pi }\cos 0} \right)\\& = e - e - \frac{4}{\pi }\left( {\frac{1}{{\sqrt 2 }}} \right) + 1 + \frac{4}{\pi } = 1 + \frac{4}{\pi } - \frac{{2\sqrt 2 }}{\pi }\end{align}\]

Chapter 7 Ex.7.9 Question 21

\(\begin{align}&\int_1^{\sqrt 3 } {\frac{{dx}}{{1 + {x^2}}}} \\&A.\,\,\,\frac{\pi }{3}\\&B.\,\,\,\frac{{2\pi }}{3}\\&C.\,\,\,\frac{\pi }{6}\\&D.\,\,\,\frac{\pi }{{12}}\,\,\,\end{align}\)

Solution

\(\int {\frac{{dx}}{{1 + {x^2}}} = {{\tan }^{ - 1}}x} = F\left( x \right)\)

Using second fundamental theorem of calculus, we get

\[\begin{align}\int_1^{\sqrt 3 } {\frac{{dx}}{{1 + {x^2}}}}& = F\left( {\sqrt 3 } \right) - F\left( 1 \right)\\ &= {\tan ^{ - 1}}\sqrt 3 - {\tan ^{ - 1}}1\\ &= \frac{\pi }{3} - \frac{\pi }{4}\\ &= \frac{\pi }{{12}}\end{align}\]

Thus, the correct option is D.

Chapter 7 Ex.7.9 Question 22

\(\begin{align}&\int_0^{\frac{2}{3}} {\frac{{dx}}{{4 + 9{x^2}}}} \\&A.\,\,\,\frac{\pi }{6}\\&B.\,\,\,\frac{\pi }{{12}}\\&C.\,\,\,\frac{\pi }{{24}}\\&D.\,\,\,\frac{\pi }{4}\,\,\,\,\,\, \end{align}\)

Solution

\(\int {\frac{{dx}}{{4 + 9{x^2}}}} = \int {\frac{{dx}}{{{{\left( 2 \right)}^2} + {{\left( {3x} \right)}^2}}}} \)

Put \(3x = t \Rightarrow 3\;dx = dt\)

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

\[\begin{align} \qquad &= \frac{1}{3}\left[ {\frac{1}{2}{{\tan }^{ - 1}}\frac{t}{2}} \right]\\ &= \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{3x}}{2}} \right)\\&= F\left( x \right)\end{align}\]

Using second fundamental theorem of calculus, we get

\[\begin{align}\int_0^{\frac{2}{3}} {\frac{{dx}}{{4 + 9{x^2}}}} &= F\left( {\frac{2}{3}} \right) - F\left( 0 \right)\\& = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{3}{2}.\frac{2}{3}} \right) - \frac{1}{6}{\tan ^{ - 1}}0\\ &= \frac{1}{6}{\tan ^{ - 1}}1 - 0\\ &= \frac{1}{6} \times \frac{\pi }{4}\\ &= \frac{\pi }{{24}}\end{align}\]

Thus, the correct option is C.

  
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