# NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.11

## Chapter 7 Ex.7.11 Question 1

By using the properties of definite integrals, evaluate the integral $$\int_0^{\frac{\pi }{2}} {{{\cos }^2}xdx}$$

### Solution

\begin{align}I &= \int_0^{\frac{\pi }{2}} {{{\cos }^2}xdx\,\,\,\,\,\,\, \ldots \left( 1 \right)} \\ &\Rightarrow \;I = \int_0^{\frac{\pi }{2}} {{{\cos }^2}\left( {\frac{\pi }{2} - x} \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^{\frac{\pi }{2}} {{{\sin }^2}xdx} \,\,\,\,\,\,\,\, \ldots \left( 2 \right)\end{align}

Adding ($$1$$) and ($$2$$), we get

\begin{align}2I &= \int_0^{\frac{\pi }{2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)dx} \\& \Rightarrow \;2I = \int_0^{\frac{\pi }{2}} {1.dx} \\ &\Rightarrow \;2I = \left[ x \right]_0^{\frac{\pi }{2}}\\ &\Rightarrow \;2I = \frac{\pi }{2}\\& \Rightarrow \;I = \frac{\pi }{4}\end{align}

## Chapter 7 Ex.7.11 Question 2

By using the properties of definite integrals, evaluate the integral $$\int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx}$$

### Solution

$$\int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx}$$

Consider,$$I = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx} \,\,\,\, \ldots \left( 1 \right)$$

\begin{align} &\Rightarrow \;I = I = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} + \sqrt {\cos \left( {\frac{\pi }{2} - x} \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^{\frac{\pi }{2}} {\frac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}} dx\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\end{align}

Adding ($$1$$) and ($$2$$), we get

\begin{align}2I &= \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\sin x} + \sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx\\ &\Rightarrow \;\; \; 2I = \int_0^{\frac{\pi }{2}} {1.dx} \\ &\Rightarrow \;\; \; 2I = \left[ x \right]_0^{\frac{\pi }{2}}\\ &\Rightarrow \;\; \; 2I = \frac{\pi }{2}\\& \Rightarrow \;\; \; I = \frac{\pi }{4}\end{align}

## Chapter 7 Ex.7.11 Question 3

By using the properties of definite integrals, evaluate the integral $$\int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^{\frac{3}{2}}}xdx}}{{{{\sin }^{\frac{3}{2}}}x + {{\cos }^{\frac{3}{2}}}x}}dx}$$

### Solution

Let $$I = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^{\frac{3}{2}}}xdx}}{{{{\sin }^{\frac{3}{2}}}x + {{\cos }^{\frac{3}{2}}}x}}dx} \,\,\,\,\, \ldots \left( 1 \right)$$

\begin{align} &\Rightarrow \;\; \; I = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^{\frac{3}{2}}}\left( {\frac{\pi }{2} - x} \right)dx}}{{{{\sin }^{\frac{3}{2}}}\left( {\frac{\pi }{2} - x} \right) + {{\cos }^{\frac{3}{2}}}\left( {\frac{\pi }{2} - x} \right)}}dx} \,\,\,\,\,\,\left( {\int_0^a {f\left( x \right)} = \int_0^a {f\left( {a - x} \right)dx} } \right)\\ &\Rightarrow \;\; \; I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{\frac{3}{2}}}xdx}}{{{{\cos }^{\frac{3}{2}}}x + {{\sin }^{\frac{3}{2}}}x}}dx} \,\,\,\, \ldots \left( 3 \right)\end{align}

Adding ($$1$$) and ($$2$$), we get

\begin{align}2I& = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^{\frac{3}{2}}}x + {{\cos }^{\frac{3}{2}}}x}}{{{{\sin }^{\frac{3}{2}}}x + {{\cos }^{\frac{3}{2}}}x}}dx} \\ &\Rightarrow \;\; \; 2I = \int_0^{\frac{\pi }{2}} {1.dx} \Rightarrow \;\; \; 2I = \left[ x \right]_0^{\frac{\pi }{2}}\\ &\Rightarrow \;\; \; 2I = \frac{\pi }{2} \Rightarrow \;\; \; I = \frac{\pi }{4}\end{align}

## Chapter 7 Ex.7.11 Question 4

By using the properties of definite integrals, evaluate the integral $$\int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}xdx}}{{{{\sin }^5}x + {{\cos }^5}x}}dx}$$

### Solution

Consider,

\begin{align}I &= \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}xdx}}{{{{\sin }^5}x + {{\cos }^5}x}}dx} \,\,\, \ldots \left( 1 \right)\\& \Rightarrow \;\; \; I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}\left( {\frac{\pi }{2} - x} \right)dx}}{{{{\sin }^5}\left( {\frac{\pi }{2} - x} \right) + {{\cos }^5}\left( {\frac{\pi }{2} - x} \right)}}dx} \,\,\,\,\,\,\,\,\,\left( {\int_0^a {f\left( x \right)} = \int_0^a {f\left( {a - x} \right)dx} } \right)\\& \Rightarrow \;\; \; I = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}}dx} \,\,\,\,\, \ldots \left( 2 \right)\end{align}

Adding ($$1$$) and ($$2$$), we get

\begin{align}2I &= \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^5}x + {{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}} dx\\& \Rightarrow \;\; \; 2I = \int_0^{\frac{\pi }{2}} {1.dx} \Rightarrow \;\; \; 2I = \left[ x \right]_0^{\frac{\pi }{2}}\\& \Rightarrow \;\; \; 2I = \frac{\pi }{2} \Rightarrow \;\; \; I = \frac{\pi }{4}\end{align}

## Chapter 7 Ex.7.11 Question 5

By using the properties of definite integrals, evaluate the integral $$\int_{ - 5}^5 {\left| {x + 2} \right|dx}$$

### Solution

Let $$I = \int_{ - 5}^5 {\left| {x + 2} \right|dx}$$

As, $$\left( {x + 2} \right) \le 0$$ on $$\left[ { - 5, - 2} \right]$$ and $$\left( {x + 2} \right) \ge 0$$on $$\left[ { - 2,5} \right]$$

\begin{align}&\therefore \;\int_{ - 5}^5 {\left| {x + 2} \right|dx} = \int_{ - 5}^{ - 2} { - \left( {x + 2} \right)dx + } \int_{ - 2}^5 {\left( {x + 2} \right)dx} \,\,\,\,\,\;\;\;\;\;\left( {\int_a^b {f\left( x \right)} = \int_a^c {f\left( x \right)} + \int_c^b {f\left( x \right)} } \right)\\I& = - \left[ {\frac{{{x^2}}}{2} + 2x} \right]_{ - 5}^{ - 2} + \left[ {\frac{{{x^2}}}{2} + 2x} \right]_{ - 2}^5\\ &= - \left[ {\frac{{{{\left( { - 2} \right)}^2}}}{2} + 2\left( { - 2} \right) - \frac{{{{\left( { - 5} \right)}^2}}}{2} - 2\left( { - 5} \right)} \right] + \left[ {\frac{{{{\left( 5 \right)}^2}}}{2} + 2\left( 5 \right) - \frac{{{{\left( { - 2} \right)}^2}}}{2} - 2\left( { - 2} \right)} \right]\\& = - \left[ {2 - 4 - \frac{{25}}{2} + 10} \right] + \left[ {\frac{{25}}{2} + 10 - 2 + 4} \right]\\ &= - 2 + 4 + \frac{{25}}{2} - 10 + \frac{{25}}{2} + 10 - 2 + 4\\& = 29\end{align}

## Chapter 7 Ex.7.11 Question 6

By using the properties of definite integrals, evaluate the integral $$\int_2^8 {\left| {x - 5} \right|dx}$$

### Solution

Consider, $$I = \int_2^8 {\left| {x - 5} \right|dx}$$

As $$\left( {x - 5} \right) \le 0\,\,{\text{on }}\left[ {2,5} \right]\,\,{\text{and }}\left( {x - 5} \right) \ge 0\,\,{\text{on }}\left[ {5,8} \right]$$

\begin{align}I &= \int_2^5 { - \left( {x - 5} \right)dx} + \int_2^8 {\left( {x - 5} \right)dx} \,\,\,\,\,\,\,\left( {\int_a^b {f\left( x \right)} = \int_a^c {f\left( x \right)} + \int_c^b {f\left( x \right)} } \right)\\ &= - \left[ {\frac{{{x^2}}}{2} - 5x} \right]_2^5 + \left[ {\frac{{{x^2}}}{2} - 5x} \right]_5^8\\& = - \left[ {\frac{{25}}{2} - 25 - 2 + 10} \right] + \left[ {32 - 40 - \frac{{25}}{2} + 25} \right] = 9\end{align}

## Chapter 7 Ex.7.11 Question 7

By using the properties of definite integrals, evaluate the integral $$\int_0^1 {x{{\left( {1 - x} \right)}^n}dx}$$

### Solution

Consider, $$I = \int_0^1 {x{{\left( {1 - x} \right)}^n}dx}$$

\begin{align}\therefore I &= \int_0^1 {\left( {1 - x} \right){{\left( {1 - \left( {1 - x} \right)} \right)}^n}dx} \\ &= \int_0^1 {\left( {1 - x} \right){{\left( x \right)}^n}} dx = \int_0^1 {\left( {{x^n} - {x^{n + 1}}} \right)} dx\\ &= \left[ {\frac{{{x^{n + 1}}}}{{n + 1}} - \frac{{{x^{n + 2}}}}{{n + 2}}} \right]_0^1\,\,\,\,\,\,\,\,\,\,\left( {\int_1^a {f\left( x \right)} dx = \int_0^a {f\left( {a - x} \right)dx} } \right)\\ &= \left[ {\frac{1}{{n + 1}} - \frac{1}{{n + 2}}} \right]\\& = \frac{{\left( {n + 2} \right) - \left( {n + 1} \right)}}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\\ &= \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\end{align}

## Chapter 7 Ex.7.11 Question 8

By using the properties of definite integrals, evaluate the integral $$\int_0^{\frac{\pi }{4}} {\log \left( {1 + \tan x} \right)dx}$$

### Solution

Let $$I = \int_0^{\frac{\pi }{4}} {\log \left( {1 + \tan x} \right)dx} \,\,\,\,\, \ldots \left( 1 \right)$$

\begin{align}\therefore I &= I = \int_0^{\frac{\pi }{4}} {\log \left( {1 + \tan \left( {\frac{\pi }{4} - 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^{\frac{\pi }{4}} {\log \left\{ {1 + \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}}} \right\}dx} \,\,\,\,\,\;\;\left( {\tan \left( {a - b} \right) = \frac{{\tan a - \tan b}}{{1 + \tan a\tan b}}} \right)\\& \Rightarrow \;\; \; I = \int_0^{\frac{\pi }{4}} {\log \left\{ {1 + \frac{{1 - \tan x}}{{1 + \tan x}}} \right\}} dx \Rightarrow \;\; \; I = \int_0^{\frac{\pi }{4}} {\log \frac{2}{{\left( {1 + \tan x} \right)}}} dx\\ &\Rightarrow \;\; \; I = \int_0^{\frac{\pi }{4}} {\log 2dx - \int_0^{\frac{\pi }{4}} {\log \left( {1 + \tan x} \right)dx} } \\& \Rightarrow \;\; \; I = \int_0^{\frac{\pi }{4}} {\log 2dx - I} \,\,\,\,\left[ {{\text{from}}\left( 1 \right)} \right]\\ &\Rightarrow \;\; \; 2I = \log 2\left[ x \right]_0^{\frac{\pi }{4}}\\& \Rightarrow \;\; \; 2I = \log 2\left[ {\frac{\pi }{4} - 0} \right]\\I& = \frac{\pi }{8}\log 2\end{align}

## Chapter 7 Ex.7.11 Question 9

By using the properties of definite integrals, evaluate the integral $$\int_0^2 {x\sqrt {2 - x} } dx$$

### Solution

Consider, $$I = \int_0^2 {x\sqrt {2 - x} } dx$$

\begin{align}I &= \int_0^2 {\left( {2 - x} \right)\sqrt {2 - \left( {2 - x} \right)} } dx\,\,\,\,\,\left( {\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} } \right)\\ &= \int_0^2 {\left( {2 - x} \right)\sqrt x } dx\\ &= \int_0^2 {\left\{ {2{x^{\frac{1}{2}}} - {x^{\frac{3}{2}}}} \right\}} dx = \left[ {2\left( {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right) - \frac{{{x^{\frac{5}{2}}}}}{{\frac{5}{2}}}} \right]_0^2\\&= \left[ {\frac{4}{3}{x^{\frac{3}{2}}} - \frac{2}{5}{x^{\frac{5}{2}}}} \right]_0^2 = \frac{4}{3}{\left( 2 \right)^{\frac{3}{2}}} - \frac{2}{5}{\left( 2 \right)^{\frac{5}{2}}}\\ &= \frac{{4 \times 2\sqrt 2 }}{3} - \frac{2}{5} \times 4\sqrt 2 = \frac{{8\sqrt 2 }}{3} - \frac{{8\sqrt 2 }}{5}\\& = \frac{{40\sqrt 2 - 24\sqrt 2 }}{{15}} = \frac{{16\sqrt 2 }}{{15}}\end{align}

## Chapter 7 Ex.7.11 Question 10

By using the properties of definite integrals, evaluate the integral $$\int_0^{\frac{\pi }{2}} {\left( {2\log \sin x - \log \sin 2x} \right)dx}$$

### Solution

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

\begin{align}I &= \int_0^{\frac{\pi }{2}} {\left( {2\log \sin x - \log \left( {2\sin x\cos x} \right)} \right)dx} \\I &= \int_0^{\frac{\pi }{2}} {\left( {2\log \sin x - \log \sin x - \log \cos x - \log 2} \right)dx} \\ \Rightarrow \;\; \; I &= \int_0^{\frac{\pi }{2}} {\left\{ {\log \sin x - \log \cos x - \log 2} \right\}} \,\,\,\, \ldots \left( 1 \right)\end{align}

Since, $$\left( {\int_0^a {f\left( x \right)dx = \int_0^a {f\left( {a - x} \right)dx} } } \right)$$

$$\Rightarrow \;\; \; I = \int_0^{\frac{\pi }{2}} {\left\{ {\log \cos x - \log \sin x - \log 2} \right\}dx} \,\,\,\,\, \ldots \left( 2 \right)$$

Adding ($$1$$) and ($$2$$), we get

\begin{align}2I &= \int_0^{\frac{\pi }{2}} {\left( { - \log 2 - \log 2} \right)dx} \\& \Rightarrow \;\; \; 2I = - 2\log 2\int_0^{\frac{\pi }{2}} {1.dx} \\ &\Rightarrow \;\; \; I = - \log 2\left[ {\frac{\pi }{2}} \right]\\& \Rightarrow \;\; \; I = \frac{\pi }{2}\left( { - \log 2} \right)\\& \Rightarrow \;\; \; I = \frac{\pi }{2}\left[ {\log \frac{1}{2}} \right]\\ &\Rightarrow \;\; \; I = \frac{\pi }{2}\log \frac{1}{2}\end{align}

## Chapter 7 Ex.7.11 Question 11

By using the properties of definite integrals, evaluate the integral $$\int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^2}xdx}$$

### Solution

Let $$I = \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^2}xdx}$$

As $${\sin ^2}\left( { - x} \right) = {\left( {\sin \left( { - x} \right)} \right)^2} = {\left( { - \sin x} \right)^2} = {\sin ^2}x$$, therefore $${\sin ^2}x$$is an even function.

If $$f\left( x \right)$$is an even function, then $$\int_{ - a}^a {f\left( x \right)dx} = 2\int_0^a {f\left( x \right)} dx$$

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

## Chapter 7 Ex.7.11 Question 12

By using the properties of definite integrals, evaluate the integral $$\int_0^\pi {\frac{{xdx}}{{1 + \sin x}}}$$

### Solution

Let $$I = \int_0^\pi {\frac{{xdx}}{{1 + \sin x}}} \,\,\,\,\,\,\,\, \ldots \left( 1 \right)$$

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

Adding (1) and (2), we get

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

## Chapter 7 Ex.7.11 Question 13

By using the properties of definite integrals, evaluate the integral $$\int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^7}x} dx$$

### Solution

Let $$I = \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^7}x} dx\,\,\,\, \ldots \left( 1 \right)$$

As $${\sin ^7}\left( { - x} \right) = {\left( {\sin \left( { - x} \right)} \right)^7} = {\left( { - \sin x} \right)^7} = - {\sin ^7}x$$, thus $${\sin ^2}x$$is an odd function.

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

$$\therefore I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\sin }^7}xdx} = 0$$

## Chapter 7 Ex.7.11 Question 14

By using the properties of definite integrals, evaluate the integral $$\int_0^{2\pi } {{{\cos }^5}xdx}$$

### Solution

Let $$I = \int_0^{2\pi } {{{\cos }^5}xdx} \,\,\, \ldots \left( 1 \right)$$

$${\cos ^5}\left( {2\pi - x} \right) = {\cos ^5}x$$

We know that,

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

$$= 0$$ if $$f\left( {2a - x} \right) = - f\left( x \right)$$

\begin{align}\therefore I &= 2\int_0^{2\pi } {{{\cos }^5}xdx} \\& \Rightarrow \;\; \; I = 2\left( 0 \right) = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{{\cos }^5}\left( {\pi - x} \right) = - {{\cos }^5}x} \right]\end{align}

## Chapter 7 Ex.7.11 Question 15

By using the properties of definite integrals, evaluate the integral $$\int_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}}$$

### Solution

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

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

Adding ($$1$$) and ($$2$$), we get

$$\Rightarrow \;\; \; 2I = \int_0^{\frac{\pi }{2}} {\frac{0}{{1 + \sin x\cos x}}} dx \Rightarrow \;\; \; I = 0$$

## Chapter 7 Ex.7.11 Question 16

By using the properties of definite integrals, evaluate the integral $$\int_0^\pi {\log \left( {1 + \cos x} \right)} dx$$

### Solution

Consider, $$I = \int_0^\pi {\log \left( {1 + \cos x} \right)} dx\,\,\,\,\,\,\, \ldots \left( 1 \right)$$

\begin{align} &\Rightarrow \;\; \; I = \int_0^\pi {\log \left( {1 + \cos \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 {\log \left( {1 - \cos x} \right)dx} \,\,\,\,\,\, \ldots \left( 2 \right)\end{align}

Adding ($$1$$) and ($$2$$), we get

\begin{align}2I &= \int_0^\pi {\left\{ {\log \left( {1 + \cos x} \right) + \log \left( {1 - \cos x} \right)} \right\}dx} \\& \Rightarrow \;\; \; 2I = \int_0^\pi {\log \left( {1 - {{\cos }^2}x} \right)dx} \\ &\Rightarrow \;\; \; 2I = \int_0^\pi {\log \left( {{{\sin }^2}x} \right)dx} \\ &\Rightarrow \;\; \; 2I = 2\int_0^\pi {\log \left( {\sin x} \right)dx} \\& \Rightarrow \;\; \; I = \int_0^\pi {\log \left( {\sin x} \right)dx} \,\,\,\,\,\,\,\,\,\, \ldots \left( 3 \right)\\&\therefore {\text{ }}\sin \left( {\pi - x} \right) = \sin x\\&{\text{We know that}},\\&\int_0^{2a} {f\left( x \right)dx} = 2\int_0^a {f\left( x \right)} dx{\text{ if }}f\left( {2a - x} \right) = f\left( x \right)\\&\therefore I = \int_0^{\frac{\pi }{2}} {\log \sin xdx} \,\,\,\,\,\, \ldots \left( 4 \right)\\& \Rightarrow \;\; \; I = 2\int_0^{\frac{\pi }{2}} {\log \sin \left( {\frac{\pi }{2} - x} \right)dx = 2\int_0^{\frac{\pi }{2}} {\log \cos xdx} } \,\,\,\,\,\,\,\,\, \ldots \left( 5 \right)\end{align}

Adding ($$4$$) and ($$5$$), we get

\begin{align}2I &= \int_0^{\frac{\pi }{2}} {\left( {\log \sin x + \log \cos xdx} \right)} \\& \Rightarrow \;\; \; I = \int_0^{\frac{\pi }{2}} {\left( {\log \sin x + \log \cos x + \log 2 - \log 2} \right)dx} \\& \Rightarrow \;\; \; I = \int_0^{\frac{\pi }{2}} {\left( {\log 2\sin x\cos x - \log 2} \right)} dx\\ &\Rightarrow \;\; \; I = \int_0^{\frac{\pi }{2}} {\log \sin 2xdx} - \int_0^{\frac{\pi }{2}} {\log 2dx} \\&{\text{put, }}2x = t \Rightarrow \;\; \; 2dx = dt\\&{\text{When }}x = 0,t = 0\\&\therefore I = \frac{{1\pi }}{2}\int_0^a {\log \sin tdt} - \frac{\pi }{2}\log 2\\ &\Rightarrow \;\; \; I = \frac{1}{2} - \frac{\pi }{2}\log 2\\ &\Rightarrow \;\; \; I = - \pi \log 2\end{align}

## Chapter 7 Ex.7.11 Question 17

By using the properties of definite integrals, evaluate the integral $$\int_0^a {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}} dx$$

### Solution

Let $$I = \int_0^a {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}} dx$$ …($$1$$)

We know that, $$\left( {\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} } \right)$$

$$I = \int_{}^{} {\frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\, \cdots \left( 2 \right)$$

Adding ($$1$$) and ($$2$$), we get

\begin{align}2I &= \int_0^a {\frac{{\sqrt x + \sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}} dx\\ &\Rightarrow \;\; \; 2I = \int_0^a {1.dx} \\& \Rightarrow \;\; \; 2I = \left[ x \right]_0^a\\& \Rightarrow \;\; \; 2I = a\\ &\Rightarrow \;\; \; I = \frac{a}{2}\end{align}

## Chapter 7 Ex.7.11 Question 18

By using the properties of definite integrals, evaluate the integral $$\int_0^4 {\left| {x - 1} \right|dx}$$

### Solution

$$\int_0^4 {\left| {x - 1} \right|dx}$$

Since,

\begin{align}&\left( {x - 1} \right) \le 0{\text{ when 0}} \le x \le {\text{1 and}}\left( {x - 1} \right) \ge 0{\text{ when 1}} \le x \le 4\\I &= \int_0^1 {\left| {x - 1} \right|dx} + \int_1^4 {\left| {x - 1} \right|dx} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\int_b^b {f\left( x \right)dx} = \int_b^c {f\left( x \right)dx} + \int_c^b {f\left( x \right)dx} } \right)\\I &= \int_0^1 { - \left( {x - 1} \right)dx + \int_0^4 {\left( {x - 1} \right)dx} } \\ &= \left[ {x - \frac{{{x^2}}}{2}} \right]_0^1 + \left[ {\frac{{{x^2}}}{2} - x} \right]_1^4 = 1 - \frac{1}{2} + \frac{{{{\left( 4 \right)}^2}}}{2} - 4 - \frac{1}{2} + 1\\ &= 1 - \frac{1}{2} + 8 - 4 - \frac{1}{2} + 1\\ &= 5\end{align}

## Chapter 7 Ex.7.11 Question 19

Show that $$\int_0^a {f\left( x \right)g\left( x \right)dx} = 2\int_0^a {f\left( x \right)} dx$$ if f and g are defined as $$f\left( x \right) = f\left( {a - x} \right)\,\,{\text{and }}g\left( x \right) = \left( {a - x} \right) = 4$$

### Solution

Let

\begin{align}I &= \int_0^a {f\left( x \right)g\left( x \right)dx} \,\,\,\, \ldots \left( 1 \right)\\ &\Rightarrow \;\; \; \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int_0^a {f\left( x \right)dx = \int_0^a {f\left( {a - x} \right)dx} } } \right)\\ &\Rightarrow \;\; \; \int_0^a {f\left( x \right)g\left( {a - x} \right)dx} \,\,\,\,\,\,\, \ldots \left( 2 \right)\end{align}

Adding ($$1$$) and ($$2$$), we get

\begin{align}&2I = \int_0^a {\left\{ {f\left( x \right)g\left( x \right) + f\left( x \right)g\left( {a - x} \right)} \right\}dx} \\& \Rightarrow 2I = \int_0^a {f\left( x \right)\left\{ {g\left( x \right) + g\left( {a - x} \right)} \right\}dx} \\ &\Rightarrow 2I = \int_0^a {f\left( x \right) \times 4dx} \,\,\,\,\,\,\,\,\,\,\left[ {g\left( x \right) + g\left( {a - x} \right) = 4} \right]\\& \Rightarrow I = 2\int_0^a {f\left( x \right)dx} \end{align}

## Chapter 7 Ex.7.11 Question 20

The value of $$\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {{x^3} + x\cos x + {{\tan }^5}x + 1} \right)} dx$$ is

A. $$0$$

B. $$2$$

C. $$\pi$$

D. $$1$$

### Solution

Consider, $$\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {{x^3} + x\cos x + {{\tan }^5}x + 1} \right)} dx$$

$$\Rightarrow \;\; \; I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{x^3}dx} + \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {x\cos xdx} + \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\tan }^5}xdx} + \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {1.dx}$$

For $$f\left( x \right)$$ an even function, then $$\int_{ - a}^a {f\left( x \right)dx} = 2\int_0^a {f\left( x \right)dx}$$

If $$f\left( x \right)$$ is an odd function, then $$\int_{ - a}^a {f\left( x \right)dx}$$

And

$$I = 0 + 0 + 0 + 2\int_0^{\frac{\pi }{2}} {1.dx}$$

\begin{align} &= 2\left[ x \right]_0^{\frac{\pi }{2}}\\& = \frac{{2\pi }}{2}\\& = \pi\end{align}

Thus, the correct is option $$C$$.

## Chapter 7 Ex.7.11 Question 21

The value of $$\int_0^{\frac{\pi }{2}} {\left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} dx$$ is

A. $$2$$

B. $$\frac{3}{4}$$

C. $$0$$

D. $$-2$$

### Solution

Let $$I = \int_0^{\frac{\pi }{2}} {\left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} dx\,\,\,\,\,\,\,\, \ldots \left( 1 \right)$$

\begin{align} &\Rightarrow \;\; \; I = I = \int_0^{\frac{\pi }{2}} {\left( {\frac{{4 + 3\sin \left( {\frac{\pi }{2} - x} \right)}}{{4 + 3\cos \left( {\frac{\pi }{2} - 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^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\end{align}

Adding ($$1$$) and ($$2$$), we get

\begin{align}2I &= \int_0^{\frac{\pi }{2}} {\left\{ {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) + \log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \right\}dx} \\ &\Rightarrow \;\; \; 2I = \int_0^{\frac{\pi }{2}} {\left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}} \times \frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} dx\\ &\Rightarrow \;\; \; 2I = \int_0^{\frac{\pi }{2}} {\log 1dx} \\& \Rightarrow \;\; \; 2I = \int_0^{\frac{\pi }{2}} {0dx} \\&\Rightarrow \;\; \; I = 0\end{align}

Thus, the correct option is C.

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