Question

The value of intergral ${\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{x}{1+\sin x}dx$ is :

• A. $\pi \sqrt{2}$
• B. $\pi (\sqrt{2}-1)$
• C. $\frac{\pi }{2}(\sqrt{2}+1)$
• D. $2\pi (\sqrt{2}-1)$

Answer 1

The correct option is B $\pi (\sqrt{2}-1)$

Given the integral,

${\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{x}{\sin x+1}dx={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{x\frac{1}{\cos x}}{\frac{\sin x}{\cos x}+\frac{1}{\cos x}}dx={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{x\sec x}{\tan x+\sec x}dx$

Using integration by parts we get,

$=-\frac{x}{\tan x+\sec x}-{\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{1}{-\tan x-\sec x}dx$

For,

$\int \frac{1}{-\tan x-\sec x}dx=-\int \frac{1}{\tan x+\sec x}dx=-\int \cos x\frac{1}{\sin x+1}dx$

Let,

$u=\sin x\Rightarrow \frac{du}{dx}=\cos x\Rightarrow du=\cos xdx$

substituting these values we get,

$-\int \cos x\frac{1}{\sin x+1}dx=-\int \frac{1}{u+1}du$

Again let,

$v=u+1\Rightarrow \frac{dv}{du}=1\Rightarrow dv=du$

Substituting these values we get,

$\int \frac{1}{u+1}du=\int \frac{1}{v}dv=\ln \left(v\right)=\ln (u+1)=\ln (\sin x+1)\therefore \int \frac{1}{-\tan x-\sec x}dx=-\ln (\sin x+1)$

So,

$-\frac{x}{\tan x+\sec x}-\int \frac{1}{-\tan x-\sec x}dx=\ln (\sin x+1)-\frac{x}{\tan x+\sec x}\therefore {\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{x}{\sin x+1}dx={[\ln (\sin x+1)-\frac{x}{\tan x+\sec x}]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}=\ln (\sin \frac{3\pi }{4}+1)-\ln (\sin \frac{\pi }{4}+1)-[\frac{\frac{3\pi }{4}}{\tan \frac{3\pi }{4}+\sec \frac{3\pi }{4}}-\frac{\frac{\pi }{4}}{\tan \frac{\pi }{4}+\sec \frac{\pi }{4}}]=\pi (\sqrt{2}-1).$