Maths Integrals

Kindly go through the solution

$\frac{1}{2}.$

So, $\frac{1}{x-x^3}=\frac{1}{x}+\frac{1}{2}\frac{1}{\left(1-x\right)}-\frac{1}{2}\frac{1}{\left(1+x\right)}$

$\therefore {\displaystyle \int \frac{dx}{x-x^3}}={\displaystyle \int \frac{dx}{x}}+\frac{1}{2}{\displaystyle \int \frac{dx}{1-x}}-\frac{1}{2}{\displaystyle \int \frac{dx}{1+x}}$

= log |x| + $\frac{1}{2}$ log $\frac{log|1-x|}{(-1)}-\frac{1}{2}$ log |1 + x|

= $\frac{1}{2}\left[2log\left|x\right|-log|1-x|-log|1+x|\right]+\; C$

= $\frac{1}{2}\left[log{x}^2-log(1-x)(1+x)\right]+\; C$

= $\frac{1}{2}log\left(\frac{x^2}{1-x^2}\right)+\; C$

Let I = ${\displaystyle {\int }_0^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{4+3sinx}{4+3cosx}dx.$ ______(i)

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{4+3sin\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-x\right)}{4+3cos\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-x\right)}dx.$

I = ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{4+3cosx}{4+3sinx}dx$ ______(ii)

Adding (i) & (ii) we get,

2I = ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[log\left(\frac{4+3sinx}{4+3cosx}\right)+log\left(\frac{4+3cosx}{4+3sinx}\right)\right]}dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\left(\frac{4+3sinx}{4+3cosx}*\frac{4+3sinx}{4+3cosinx}\right)dx$

= ${{\displaystyle \int }}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ log 1 dx

= ${{\displaystyle \int }}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ * 0 dx

= 0

I = 0.

?Option (C) is correct

Kindly go through the solution

Given, f(x) = f(a – x)

g(x) + g(a – x) = 4.

Let I = ${\displaystyle {\int }_0^{a}f}(x)g(x)dx.$ ____(1)

= ${{\displaystyle \int }}^a$ f(x) g (a – x)dx

I = ${{\displaystyle \int }}^a$ f(x) g(a – x)dx______(2)

Adding (1) and (2),

2I = ${{\displaystyle \int }}^a$ [f(x) g(x) + f(x) g(a + x)]dx

= ${{\displaystyle \int }}^a$ f(x) [g(x) + g(a + x)]dx

2I = ${{\displaystyle \int }}^a$ f(x) 4 dx

I = $\frac{4}{2}{{\displaystyle \int }}^a$ f(x)dx = 2 ${{\displaystyle \int }}^a$ f(x)dx.

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

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}$

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

I = 2 * 0 [? cos⁵(π – x) = –cos⁵x]

I = 0.