Chapter 7: Integrals

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}$                     …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}\left( \dfrac{\pi }{2}-x \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$                          …(2)

Adding equation (1) and (2),

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}=\dfrac{\pi }{4}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$                     …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}}{\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}+\sqrt{\cos \left( \dfrac{\pi }{2}-x \right)}}dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$                                         …(2)

Adding equation (1) and (2),

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}=\dfrac{\pi }{4}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}xdx}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}}$                        …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}\left( \dfrac{\pi }{2}-x \right)dx}{{{\sin }^{\dfrac{3}{2}}}\left( \dfrac{\pi }{2}-x \right)+{{\cos }^{\dfrac{3}{2}}}\left( \dfrac{\pi }{2}-x \right)}}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{\dfrac{3}{2}}}xdx}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}}$                                         …(2)

Adding equation (1) and (2),

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}xdx}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}}=\dfrac{\pi }{4}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{5}}xdx}{{{\sin }^{5}}x+{{\cos }^{5}}x}}$                        …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{5}}\left( \dfrac{\pi }{2}-x \right)dx}{{{\sin }^{5}}\left( \dfrac{\pi }{2}-x \right)+{{\cos }^{5}}\left( \dfrac{\pi }{2}-x \right)}}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{5}}xdx}{{{\sin }^{5}}x+{{\cos }^{5}}x}}$                                         …(2)

Adding equation (1) and (2),

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{5}}x+{{\cos }^{5}}x}{{{\sin }^{5}}x+{{\cos }^{5}}x}dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{5}}xdx}{{{\sin }^{5}}x+{{\cos }^{5}}x}}=\dfrac{\pi }{4}$

Ans:  Let $I=\int\limits_{-5}^{5}{\left| x+2 \right|}dx$ 

Since, $\left( x+2 \right)\le 0$ for interval $\left[ -5,-2 \right]$ .

Therefore, $\left( x+2 \right)\ge 0$ for interval $\left[ -2,5 \right]$.

As, $\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{c}{f\left( x \right)}dx+\int\limits_{c}^{b}{f\left( x \right)dx}$

Hence, $\int\limits_{-5}^{-2}{-\left( x+2 \right)}dx+\int\limits_{-2}^{5}{\left( x+2 \right)dx}$ .

Thus, 

$I=\int\limits_{-5}^{-2}{-\left( x+2 \right)}dx+\int\limits_{-2}^{5}{\left( x+2 \right)dx}$

$=-\left[ \dfrac{{{x}^{2}}}{2}+2x \right]_{-5}^{2}+\left[ \dfrac{{{x}^{2}}}{2}+2x \right]_{-2}^{5}$

$=-\left[ \dfrac{{{\left( 2 \right)}^{2}}}{2}+2\left( 2 \right)-\dfrac{{{\left( -5 \right)}^{2}}}{2}-2\left( -5 \right) \right]+\left[ \dfrac{{{\left( 5 \right)}^{2}}}{2}+2\left( 5 \right)-\dfrac{{{\left( -2 \right)}^{2}}}{2}-2\left( -2 \right) \right]$$=-\left[ 2-4-\dfrac{25}{2}+10 \right]+\left[ \dfrac{25}{2}+10-2+4 \right]$

$=-2+4+\dfrac{25}{2}+10+\dfrac{25}{2}+10-2+4$

$=29$

Ans: Let $I=\int\limits_{2}^{8}{\left| x-5 \right|}dx$ 

Since, $\left( x-5 \right)\le 0$ for interval $\left[ 2,5 \right]$ .

Therefore, $\left( x-5 \right)\ge 0$ for interval $\left[ 5,8 \right]$.

As, $\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{c}{f\left( x \right)}dx+\int\limits_{c}^{b}{f\left( x \right)dx}$

Hence, $\int\limits_{2}^{5}{-\left( x-5 \right)}dx+\int\limits_{5}^{8}{\left( x-5 \right)dx}$ .

Thus, 

$I=\int\limits_{2}^{5}{-\left( x-5 \right)}dx+\int\limits_{5}^{8}{\left( x-5 \right)dx}$

$=-\left[ \dfrac{{{x}^{2}}}{2}-5x \right]_{2}^{5}+\left[ \dfrac{{{x}^{2}}}{2}-5x \right]_{5}^{8}$

$=-\left[ \dfrac{{{\left( 5 \right)}^{2}}}{2}-5\left( 5 \right)-\dfrac{{{\left( 2 \right)}^{2}}}{2}+5\left( 2 \right) \right]+\left[ \dfrac{{{\left( 8 \right)}^{2}}}{2}-5\left( 8 \right)-\dfrac{{{\left( 5 \right)}^{2}}}{2}+5\left( 5 \right) \right]$

$=-\left[ \dfrac{25}{2}-25-2+10 \right]+\left[ 32-40-\dfrac{25}{2}+25 \right]$

$=-\dfrac{25}{2}+25+2-10+32-40-\dfrac{25}{2}+25$

$=9$

Ans: Let  $I=\int\limits_{0}^{1}{x{{\left( 1-x \right)}^{n}}dx}$

Thus, $I=\int\limits_{0}^{1}{\left( 1-x \right){{\left( 1-\left( 1-x \right) \right)}^{n}}dx}$

Since, $\int\limits_{1}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$.

Therefore, 

$\int\limits_{0}^{1}{\left( 1-x \right){{\left( x \right)}^{n}}}dx$

$=\int\limits_{0}^{1}{\left( {{x}^{n}}-{{x}^{n+1}} \right)dx}$

$=\left[ \dfrac{{{x}^{n+1}}}{n+1}-\dfrac{{{x}^{n+2}}}{n+2} \right]_{0}^{1}$

$=\left[ \dfrac{1}{n+1}-\dfrac{1}{n+2} \right]$

$=\dfrac{\left( n+2 \right)-\left( n+1 \right)}{\left( n+1 \right)\left( n+2 \right)}$

$=\dfrac{1}{\left( n+1 \right)\left( n+2 \right)}$

Ans:  Let  $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)dx}$

Since, $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left[ 1+\tan \left( \dfrac{\pi }{4}-x \right) \right]dx}$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left[ 1+\dfrac{\tan \dfrac{\pi }{4}-\tan x}{1+\tan \dfrac{\pi }{4}\tan x} \right]dx}$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left\{ 1+\dfrac{1-\tan x}{1+\tan x} \right\}dx}$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \dfrac{2}{1+\tan x}}dx$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx}-\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)}dx$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx}-I$

$2I=\left[ x\log 2 \right]_{0}^{\dfrac{\pi }{4}}$

$2I=\dfrac{\pi }{4}\log 2$

$I=\dfrac{\pi }{8}\log 2$

Ans: Let  $I=\int\limits_{0}^{2}{x\sqrt{2-x}dx}$

Since, $\int\limits_{0}^{a}{f\left( x \right)}dx=\int\limits_{0}^{a}{f\left( a-x \right)}dx$

Therefore, $I=\int\limits_{0}^{2}{\left( 2-x \right)}\sqrt{x}dx$

$=\int\limits_{0}^{2}{\left\{ 2{{x}^{{1}/{2}\;}}-{{x}^{{3}/{2}\;}} \right\}}dx$

$=\left[ 2\left( \dfrac{{{x}^{{3}/{2}\;}}}{{3}/{2}\;} \right)-\dfrac{{{x}^{{5}/{2}\;}}}{{5}/{2}\;} \right]_{0}^{2}$

$=\left[ \dfrac{4}{3}{{x}^{{3}/{2}\;}}-\dfrac{2}{5}{{x}^{{5}/{2}\;}} \right]_{0}^{2}$

$=\dfrac{4}{3}{{\left( 2 \right)}^{{3}/{2}\;}}-\dfrac{2}{5}{{\left( 2 \right)}^{{5}/{2}\;}}$

$=\dfrac{8\sqrt{2}}{3}-\dfrac{8\sqrt{2}}{5}$

$=\dfrac{40\sqrt{2}-24\sqrt{2}}{15}$

$=\dfrac{16\sqrt{2}}{15}$

Ans: Let  $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\log \sin x-\log \sin 2x \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\log \sin x-\log \left( 2\sin x\cos x \right) \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\log \sin x-\log \sin x-\log \cos x-\log 2 \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log \sin x-\log \cos x-\log 2 \right)dx}$                      …….(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log \cos x-\log \sin x-\log 2 \right)dx}$ …….(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( -\log 2-\log 2 \right)dx}$

$\Rightarrow 2I=-2\log 2\int\limits_{0}^{\dfrac{\pi }{2}}{dx}$

$\Rightarrow I=-\log 2\left[ \dfrac{\pi }{2} \right]$

$\Rightarrow I=-\dfrac{\pi }{2}\left[ \log 2 \right]$

$\Rightarrow I=\dfrac{\pi }{2}\left[ -\log 2 \right]$

$\Rightarrow I=\dfrac{\pi }{2}\left[ \log \dfrac{1}{2} \right]$

$\Rightarrow I=\dfrac{\pi }{2}\log \dfrac{1}{2}$