Question

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

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

Answer 1

The correct option is A $\frac{\pi }{2}(\sqrt{2}+1)$

We have,

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

Multiply and divide LHS with $(1-\sin x)$, we get

$I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{x(1-\sin x)}{1-{\sin }^{2}x}dx$

$I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{x(1-\sin x)}{{\cos }^{2}x}dx$

$I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}x({\sec }^{2}x-\sec x\tan x)dx$

$I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}x{\sec }^{2}x-{\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}x\sec x\tan xdx$

Using integration by parts

$I=\left[\right(x\tan x)-(\int 1.\tan xdx){]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}$$+\left[\right(x\sec x)-(\int 1.\sec xdx){]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}$

$I=\left[\right(x\tan x)-(\log |\sec x|){]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}$$+\left[\right(x\sec x)+(\log |secx+\tan x|{]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}$

$I=\{[\frac{3\pi }{4}\tan \frac{3\pi }{4}-\ln \left|\frac{3\pi }{4}\right|]-[\frac{3\pi }{4}\sec \frac{3\pi }{4}-\ln |\sec \frac{3\pi }{4}+\tan \frac{3\pi }{4}|]\}$

$-\{[\frac{\pi }{4}\tan \frac{\pi }{4}-\ln \left|\frac{\pi }{4}\right|]-[\frac{\pi }{4}\sec \frac{\pi }{4}-\ln |\sec \frac{\pi }{4}+\tan \frac{\pi }{4}|]\}$

$I=\frac{\pi }{2}(\sqrt{2}+1)$