Integrals

$\begin{array}{l}LetI={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}log}(1+tanx)dx-----(i)\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}log}\left[1+tan\left(\frac{\pi }{4}-x\right)\right]dx?\{{\displaystyle {\int }_0^{a}f(x)dx=}{\displaystyle {\int }_0^{a}f(a-x)dx}\\ ={\displaystyle {\int }_0^{\frac{\pi }{4}}log}\left[1+\frac{tan\frac{\pi }{4}-tanx}{1+tan\frac{\pi }{4}tanx}\right]dx\\ ={\displaystyle {\int }_0^{\frac{\pi }{4}}log}·\left\{1+\frac{1-tanx}{1+tanx}\right\}dx\\ ={\displaystyle {\int }_0^{\frac{\pi }{4}}log}\left\{\frac{1+tanx+1-tanx}{1+tanx}\right\}dx\\ I={\displaystyle {\int }_0^{\frac{\pi }{4}}log}\left(\frac{2}{1+tanx}\right)dx-----(ii)\\ Adding\left(i\right)\&\left(ii\right)weget,\\ I+I={\displaystyle {\int }_0^{\frac{\pi }{4}}\left\{log(1+tanx)+log\left(\frac{2}{1+tanx}\right)\right\}}dx\\ I={\displaystyle {\int }_0^{\frac{\pi }{4}}log}\left\{\underset{\_}{1+tanx}\times \frac{2}{1+tanx}\right\}dx\\ ={\displaystyle {\int }_0^{\frac{\pi }{4}}log}2dx\\ ={[xlog2]}_0^{\frac{\pi }{4}}\\ =log2\left[\frac{\pi }{4}-0\right]=\frac{\pi }{4}log2\\ I=\frac{\pi }{8}log2.\end{array}$

$\begin{array}{l}Let\text{?}\text{?}\text{?}\text{?}I={\displaystyle {?}_0^{\raisebox{1ex}{$?$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}log}(1+tanx)dx?????(i)\\ ={\displaystyle {?}_0^{\raisebox{1ex}{$?$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}log}\left[1+tan\left(\frac{?}{4}?x\right)\right]dx?\{{\displaystyle {?}_0^{a}f(x)dx=}{\displaystyle {?}_0^{a}f(a?x)dx}\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}\left[1+\frac{tan\frac{?}{4}?tanx}{1+tan\frac{?}{4}tanx}\right]dx\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}·\left\{1+\frac{1?tanx}{1+tanx}\right\}dx\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}\left\{\frac{1+tanx+1?tanx}{1+tanx}\right\}dx\\ I={\displaystyle {?}_0^{\frac{?}{4}}log}\left(\frac{2}{1+tanx}\right)dx?????(ii)\\ Adding\left(i\right)\&\left(ii\right)weget,\\ I+I={\displaystyle {?}_0^{\frac{?}{4}}\left\{log(1+tanx)+log\left(\frac{2}{1+tanx}\right)\right\}}dx\\ I={\displaystyle {?}_0^{\frac{?}{4}}log}\left\{\underset{\_}{1+tanx}\times \frac{2}{1+tanx}\right\}dx\\ ={\displaystyle {?}_0^{\frac{?}{4}}log}2dx\\ ={[xlog2]}_0^{\frac{?}{4}}\\ =log2\left[\frac{?}{4}?0\right]=\frac{?}{4}log2\\ I=\frac{?}{8}log2.\end{array}$

$\begin{array}{l}LetI={\displaystyle {\int }_0^{1}x}\left(1-x^n\right)dx\\ ={\displaystyle {\int }_0^1(}1-x)\left[1-{\left(1-x\right)}^n\right]dx\left\{{\displaystyle {\int }_0^{a}f}(x)dx={\displaystyle {\int }_0^{a}f}(a-x)dx\right\}\\ ={\displaystyle {\int }_0^1(}1-x){(1-1+x)}^{n}dx\end{array}$

$\begin{array}{l}={\displaystyle {\int }_0^1(}1-x)x^{n}dx\\ ={\displaystyle {\int }_0^1\left(x^n-x^{n+1}\right)}dx\\ ={\left[\frac{x^{n+1}}{n+1}\right]}_0^1-{\left[\frac{x^{n+1+1}}{n+1+1}\right]}_0^1\\ =\left[\frac{1^{n+1}}{n+1}-\frac{0^{n+1}}{n+1}\right]-\left[\frac{1^{n+2}}{n+2}-\frac{0^{n+2}}{n+2}\right]\\ =\frac{1}{n+1}-\frac{1}{n+2}=\frac{(n+2)-(n+1)}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)}\end{array}$

$\begin{array}{l}LetI={\displaystyle {\int }_2^8|}x-5|dx.\left\{\begin{array}{l}x–5ifx–5>0,x>5\\ –\left(x–5\right)ifx–5<0,x<5\end{array}\right\}\\ ={\displaystyle {\int }_2^5-}(x-5)dx+{\displaystyle {\int }_5^8(}x-5)dx\\ =-{\left[\frac{x^2}{2}-5x\right]}_2^5+{\left[\frac{x^2}{2}-5x\right]}_5^8\\ =-\left[\left(\frac{5^2}{2}-5\times 5\right)-\left(\frac{2^2}{2}-5\times 2\right)\right]+\left[\left(\frac{8^2}{2}-5\times 8\right)-\left(\frac{5^2}{2}-5\times 5\right)\right]\\ =-\left[\frac{25}{2}-25-2+10\right]+[\left(32-40-\frac{25}{2}+25\right].\\ =-\frac{25}{2}+17+17-\frac{25}{2}\\ \begin{array}{ll}= 34 – 25\hfill & = 9\hfill \end{array}\end{array}$

$\begin{array}{l}LetI={\displaystyle {\int }_{-5}^5|}x+2|dx\\ ={\displaystyle {\int }_{-5}^{-2}|}x+2|dx+{\displaystyle {\int }_{-2}^5|}x+2|dx\end{array}$

? |x + 2| = x + 2 if if x + 2 > 0 => x> –2

– (x + 2) if x + 2 < 0 => x< –2.

$\begin{array}{l}={\left[-\frac{x^2}{2}-2x\right]}_{-5}^{-2}+{\left[\frac{x^2}{2}+2x\right]}_{-2}^5\\ =-\left[\left(\frac{{(-2)}^2}{2}+2(-2)\right)-\left(\frac{{\left(-5\right)}^2}{2}+2(-5)\right)\right]+\left[\left(\frac{5^2}{2}+2\times 5\right)-\left(\frac{{(-2)}^2}{2}+2(-2)\right)\right]\\ =-\left[2-4-\frac{25}{2}+10\right]+\left[\frac{25}{2}+10-2+4\right]\\ =-8+\frac{25}{2}+\frac{25}{2}+12\\ ==–8+25+12=29.\end{array}$

$\begin{array}{l}Let,I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{5}xdx}{si{n}^{5}x+co{s}^{5}x}}-----(i)\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^5\left(\frac{\pi }{2}-x\right)}{si{n}^5\left(\frac{\pi }{2}-x\right)+co{s}^5\left(\frac{\pi }{2}-x\right)}}dx\\ I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{si{n}^{5}x}{co{s}^{5}x+si{n}^{5}x}}dx-----(ii)\end{array}$

$\begin{array}{l}Adding\left(i\right)\&\left(ii\right)weget,\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{co{s}^{5}x}{si{n}^{5}x+co{s}^{5}x}+\frac{si{n}^{5}x}{co{s}^{5}x+si{n}^{5}x}\right\}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{5}x+si{n}^{5}x}{si{n}^{5}x+co{s}^{3}x}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}d}x=\frac{\pi }{2}\\ I=\frac{\pi }{4}.\end{array}$

$\begin{array}{l}Let,I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}}dx-----(i)\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\pi }{2}-x\right)}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\pi }{2}-x\right)+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{\pi }{2}-x\right)}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}}dx-----(ii)\\ \left(i\right)+\left(ii\right)\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}+\frac{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}\right\}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}{si{n}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x+co{s}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x}\right\}}dx\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}d}x=\frac{\pi }{2}\\ I=\frac{\mathrm{\pi /4}}{}\end{array}$

Kindly go through the solution

$LetI={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}co{s}^2}xdx·-----(i)$

$\begin{array}{l}={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cos{}^2}\left(\frac{\pi }{2}-x\right)dx\{?{\displaystyle {\int }_0^{a}f}(x)={\displaystyle {\int }_0^{a}f}(a-x)dx.\\ I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}si{n}^{2}xdx}-----(ii)\\ Adding\left(i\right)\&\left(ii\right),\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(co{s}^{2}x+si{n}^{2}x\right)}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}1}·dx\\ 2I={[x]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\pi }{2}-0\\ 2I=\frac{\pi }{2}\\ I=\frac{\pi }{}\end{array}$