Mathematics - Definite Integration Question with Solution | TestHub

Given:

$f\left(x\right)=\int \frac{\mathrm{cosec}x+\sin x}{\mathrm{cosec}x\sec x+\tan x{\sin }^{2}x}dx$

$\Rightarrow f\left(x\right)=\int \frac{{\displaystyle \frac{1}{\sin x}}+\sin x}{{\displaystyle \frac{1}{\sin x\cos x}}+{\displaystyle \frac{\sin x}{\cos x}}\left({\sin }^{2}x\right)}dx$

$\Rightarrow f\left(x\right)=\int \frac{{\displaystyle \frac{1+{\sin }^{2}x}{\sin x}}}{{\displaystyle \frac{1+{\sin }^{4}x}{\sin x\cos x}}}dx$

$\Rightarrow f\left(x\right)=\int \frac{{\displaystyle \left(1+{\sin }^{2}x\right)\cos x}}{{\displaystyle 1+{\sin }^{4}x}}dx$

Let, $\sin x=t$

$\cos xdx=dt$

$\Rightarrow f\left(x\right)=\int \frac{{\displaystyle 1+t^2}}{{\displaystyle 1+t^4}}dt$

$\Rightarrow f\left(x\right)=\int \frac{{\displaystyle 1+\frac{1}{t^2}}}{{\displaystyle t^2+\frac{1}{t^2}}}dt$

$\Rightarrow f\left(x\right)=\int \frac{{\displaystyle 1+\frac{1}{t^2}}}{{\displaystyle {\left(t-\frac{1}{t}\right)}^2+2}}dt$

Let, $t-\frac{1}{t}=u$

$\Rightarrow f\left(x\right)=\int \frac{{\displaystyle 1}}{{\displaystyle u^2+2}}du$

$\Rightarrow f\left(x\right)=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)+C$

$\Rightarrow f\left(x\right)=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t-{\displaystyle \frac{1}{t}}}{\sqrt{2}}\right)+C$

$\Rightarrow f\left(x\right)=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sin x-{\displaystyle \frac{1}{\sin x}}}{\sqrt{2}}\right)+C$

$\Rightarrow f\left(x\right)=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sin x-{\displaystyle \mathrm{cosec}x}}{\sqrt{2}}\right)+C$

Now, $\underset{x\to {\frac{\pi }{2}}^{-}}{\lim }f\left(x\right)=0$

$\Rightarrow \underset{x\to {\frac{\pi }{2}}^{-}}{\lim }f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{-}}{\lim }\left[\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sin x-{\displaystyle \mathrm{cosec}x}}{\sqrt{2}}\right)+C\right]=0$

$\Rightarrow \frac{1}{\sqrt{2}}{\tan }^{-1}\left(0\right)+C=0$

$\Rightarrow C=0$

$\Rightarrow f\left(x\right)=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sin x-{\displaystyle \mathrm{cosec}x}}{\sqrt{2}}\right)$

$\Rightarrow y\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sin \frac{\pi }{4}-{\displaystyle \mathrm{cosec}\frac{\pi }{4}}}{\sqrt{2}}\right)$

$\Rightarrow y\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\frac{1}{\sqrt{2}}-\sqrt{2}}{\sqrt{2}}\right)$

$\Rightarrow y\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\frac{-1}{\sqrt{2}}}{\sqrt{2}}\right)$

$\Rightarrow y\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{-1}{2}\right)$