Question

The value of integral $\underset{\pi /4}{\overset{3\pi /4}{\int }}\frac{x}{1+\sin x}dx$ is :

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

Answer 1

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

$I=\underset{\pi /4}{\overset{3\pi /4}{\int }}\frac{x}{1+\sin x}dx...\left(1\right)$
$\Rightarrow I=\underset{\pi /4}{\overset{3\pi /4}{\int }}\frac{\pi -x}{1+\sin (\pi -x)}dx$
$\because \underset{a}{\overset{b}{\int }}f\left(x\right)=\underset{a}{\overset{b}{\int }}f(a+b-x)$
$\Rightarrow I=\underset{\pi /4}{\overset{3\pi /4}{\int }}\frac{\pi -x}{1+\sin x}dx...\left(2\right)$
Add $\left(1\right)$ and $\left(2\right)$
$\Rightarrow 2I=\underset{\pi /4}{\overset{3\pi /4}{\int }}\frac{\pi }{1+\sin x}dx$
$\Rightarrow 2I=\underset{\pi /4}{\overset{3\pi /4}{\int }}\frac{\pi (1-\sin x)}{1-{\sin }^{2}x}dx$
$\Rightarrow I=\frac{\pi }{2}\underset{\pi /4}{\overset{3\pi /4}{\int }}({\sec }^{2}x-\sec x\tan x)dx$
$\Rightarrow I=\frac{\pi }{2}[(\tan \frac{3\pi }{4}-\tan \frac{\pi }{4})-(\sec \frac{3\pi }{4}-\sec \frac{\pi }{4})]$
$\Rightarrow I=\pi (\sqrt{2}-1)$