Question

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

• A. $(\sqrt{2}-1)\pi$
• B. $(\sqrt{2}+1)\pi$
• C. $\pi$
• D. none of these

Answer 1

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

$I={\int }_{\pi /4}^{3\pi /4}\frac{x}{1+\sin x}dx=I_1\left(say\right)$

Now, using the property:-

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Here , $a+b=\pi$

$I={\int }_{\pi /4}^{3\pi /4}\frac{\pi -x}{1+\sin (\pi -x)}dx=I_2\left(say\right)$

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

$2I=I_1+I_2=\pi {\int }_{\pi /4}^{3\pi /4}\frac{1}{1+\sin x}dx$

$⟹2I=\pi {\int }_{\pi /4}^{3\pi /4}\frac{1}{1+\sin x}.\frac{1-\sin x}{1-\sin x}dx$

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

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

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

$⟹2I=\pi {[\tan x-\sec x]}_{\pi /4}^{3\pi /4}$

$⟹2I=\pi (2\sqrt{2}-2)$

$⟹I=(\sqrt{2}-1)\pi$

Hence, answer is option-(A).