Integrals

$\begin{array}{l}I=-{\displaystyle \underset{0}{\overset{1}{\int }}(x-1)dx}+{\displaystyle \underset{1}{\overset{4}{\int }}(x-1)dx}\\ =-[x^2/2-x]{}_0^1+[x^2/2-x]{}_1^4=5\end{array}$

Kindly go through the solution

$\begin{array}{l}LetI={\displaystyle {\int }_0^{\pi }log}\left(1+cosx\right)dx-----(i)\\ ={\displaystyle {\int }_0^{\pi }log}[1+cos(\pi -x)]dx\{?{\displaystyle {\int }_0^{a}f}(x)dx={\displaystyle {\int }_0^{a}f}(a-x)dx\\ ={{\displaystyle \int }}^{\pi }log\left(1–cosx\right)dx-----(ii)\\ Adding\left(i\right)\&\left(ii\right)\\ 2I={\displaystyle {\int }_0^{\pi }[log(1+cosx)+log(1-cosx)]}dx\end{array}$

$\begin{array}{l}LetI={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{sinx-cosx}{1+sinxcosx}}dx-----(i)\\ {\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{sin\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-x\right)-\left(cos\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-x\right)}{1+sin\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-x\right)cos\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-x\right)}}dx\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{cosx-sinx}{1+cosxsinx}}dx.\\ ={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-}\frac{(sinx-cosx)}{1+cosxsinx}dx-----(ii)\\ Adding\left(i\right)\&\left(ii\right)weget,\\ 2I={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{(sinx-cosx)}{1+sinxcosx}-\frac{(sinx-cosx}{1+sinxcosx}\right\}}dx\\ \Rightarrow I=0\end{array}$

$\begin{array}{l}LetI={\displaystyle {\int }_0^{2\pi }co{s}^5}xdx\\ =2{\displaystyle {\int }_0^{\pi }{(cosx)}^5}dx\\ Heref\left(x\right)=co{s}^{5}x\\ So, f(2\times \pi –x)=co{s}^5(2\pi –x)=co{s}^{5}x.\\ And {{\displaystyle \int }}^{2a}f\left(x\right)dx=0if,f\left(2a–x\right)=–f\left(x\right)\end{array}$

I = 2 × 0 [? cos⁵(π – x) = –cos⁵x]

I = 0.

$Let,I={\displaystyle {\int }_{-\raisebox{1ex}{$\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}\times (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\times \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\times \frac{\pi }{2}\\ I=-\frac{log{2}^2\times \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2}\\ =-\frac{2log2\times \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2}\\ =\frac{-\pi }{2}log2.\end{array}$

Kindly go through the solution