Question

Value of the integral ${\int }_0^{\pi }\frac{xdx}{1+\sin x}$, is

• A. $\frac{\pi }{6}$
• B. $\pi$
• C. $\frac{\pi }{15}$
• D. $4\pi$

Answer 1

The correct option is B $\pi$

Step 1: Given:

$I={\int }_0^{\pi }\frac{x}{1+\sin x}dx$ ...(1)

Step 2: To find:

Value of $I$

Step 3: Definite integral identity used:

${\int }_0^{a}f\left(x\right)={\int }_0^{a}f(a-x)$

Step 4: Solution:

Given integral can be evaluated as follows:
$I={\int }_0^{\pi }\frac{x}{1+\sin x}dx={\int }_0^{\pi }\frac{\pi -x}{1+\sin (\pi -x)}dx$ $={\int }_0^{\pi }\frac{\pi -x}{1+\sin x}dx$ ...(2)

Step 5:

Adding (1) and (2), we get

$2I={\int }_0^{\pi }\frac{x}{1+\sin x}dx+{\int }_0^{\pi }\frac{\pi -x}{1+\sin x}$

$\Rightarrow 2I={\int }_0^{\pi }\frac{\pi }{1+\sin x}dx$ ...(3)
Step 6:

Now we will evaluate integral on the RHS of equation (3),

${\int }_0^{\pi }\frac{\pi }{1+\sin x}dx={\int }_0^{\pi }\frac{\pi (1-\sin x)}{(1-\sin x)(1+\sin x)}$

$={\int }_0^{\pi }\frac{\pi (1-\sin x)}{1-{\sin }^{2}x}={\int }_0^{\pi }\frac{\pi (1-\sin x)}{{\cos }^{2}x}dx$

$={\int }_0^{\pi }\pi ({\sec }^{2}x-\sec x\tan x)dx$

Step 7:

Integrating term by term, we get
$=\pi \left({\int }_0^{\pi }\right({\sec }^{2}x)dx-{\int }_0^{\pi }(\sec x\tan x\left)dx\right)$

$=\pi \left(\right[\tan x{]}_0^{\pi }-\left[\sec x{]}_0^{\pi }\right)$
$=\pi [0-0-(-1-1\left)\right]$
$=2\pi$

Step 8:
Going back this result in equation (3), we will get
$2I={\int }_0^{\pi }\frac{x}{1+sinx}dx=2\pi$

$\Rightarrow I=\pi$

Step 9: Result:

$I=\pi$

Hence, option B.