Maths Integrals

$Let,I={\displaystyle {\int }_{-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}si{n}^7}xdx.$

Are f(x) = sin⁷x

f(–x) = sin⁷(–x) = –sin⁷x = –f(x).

i.e., odd function.

So, I = 0.

$\begin{array}{l}LetI={\displaystyle {\int }_0^{\pi }\frac{xdx}{1+sinx}}-----(i)\\ ={\displaystyle {\int }_0^{\pi }\frac{\pi -x}{1+sin(\pi -x)}}dx\left\{{\displaystyle {\int }_0^{a}f}(x)dx={\displaystyle {\int }_0^{a}f}(a-x)dx\right\}\\ ={\displaystyle {\int }_0^{\pi }\frac{\pi -x}{1+sinx}}dx-----(ii)\\ \left(i\right)+\left(ii\right),\\ 2I={\displaystyle {\int }_0^{\pi }\left(\frac{x}{1+sinx}+\frac{\pi -x}{1+sinx}\right)}dx\\ ={\displaystyle {\int }_0^{\pi }\frac{\pi }{1+sinx}}dx\end{array}$

$=2\pi {\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{1+sinx}}dx\left\{{\displaystyle {\int }_0^{2a}f}(x)dx=2{\displaystyle {\int }_0^{a}f}(x)dx\right\}$

$\begin{array}{l}=2\pi {\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{dx}{1+sin\left(\frac{\pi }{2}-x\right)}}\\ =2\pi {\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{dx}{1+cosx}}.\left\{?cos2x=2co{s}^{2}x-1\right\}\\ =2\pi {\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{dx}{2co{s}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ =\pi {\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}se{c}^2}\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.dx\end{array}$

$\begin{array}{l}=\pi {\left[\frac{tan\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =2\pi \left[tan\frac{\pi }{4}-tan0\right]\\ I=\frac{2\pi }{2}*(1-0)\\ I=\pi \end{array}$

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

$\begin{array}{l}\left(i\right)+\left(ii\right)weget,\\ ?{\displaystyle {\int }_{-a}^{a}f}(x)dx=2{\displaystyle {\int }_0^{a}f}(x)dxiff(-x)=f(x)\end{array}$

$\begin{array}{l}f\left(x\right) = si{n}^{2}x\hfill \end{array}$

$\begin{array}{ll} f\left(x\right) = si{n}^2\left(–x\right) = {\left(–1\right)}^{2}si{n}^{2}x= si{n}^{2}x\hfill & \hfill \end{array}$

$\begin{array}{l}if,f\left(x\right) =f\left(–x\right)\hfill \end{array}$

$\begin{array}{l}\left\{?{\displaystyle {\int }_0^{a}f}(x)dx={\displaystyle {\int }_0^{a}f}(a-x)dx\right\}\\ 2I=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(si{n}^{2}x+co{s}^{2}x\right)}dx\\ I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}1}dx=\frac{\pi }{2}.\end{array}$

$\begin{array}{l}LetI={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}(}2logsinx-logsin2x)dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(logsi{n}^{2}x-logsin2x\right)}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{si{n}^{2}x}{sin2x}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{si{n}^{2}x}{2sinx·cosx}dx.\end{array}$

$\begin{array}{l}I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\left(\frac{1}{2}tanx\right)dx-----(i)\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[log\frac{1}{2}tan\left(\frac{\pi }{2}-x\right)\right]}dx{\displaystyle {\int }_0^{a}f(x)dx=}{\displaystyle {\int }_0^{a}f(a-x)dx}\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(log\frac{1}{2}cotx\right)}dx-----(ii)\\ \left(i\right)+\left(ii\right)\\ I+I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[log\left(\frac{1}{2}tanx\right)+log\left(\frac{1}{2}cotx\right)\right]}dx\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\left(\frac{1}{2}tanx*\frac{1}{2}cotx\right)dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{1}{4}dx.\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}(}log1-log4)dx\\ =0–log4{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}d}x\\ =–log4*\frac{\pi }{2}\\ I=-\frac{log{2}^2*\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2}\\ =-\frac{2log2*\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2}\\ =\frac{-\pi }{2}log2.\end{array}$

Kindly go through the solution

$\begin{array}{l}LetI={\displaystyle {\int }_0^{\raisebox{1ex}{$\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}*\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}*\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}$

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

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

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