Question

The value of integral ${\int }_{\pi /4}^{3\pi /4}\frac{x}{1+\sin x}dx$ 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 $\pi (\sqrt{2}-1)$

$\begin{array}{c}I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{x}{1+\sin x}dx\\ I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{\left(\pi -x\right)}{1+\sin x}dx\\ 2I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{\pi }{1+\sin x}dx\\ I=\frac{\pi }{2}{\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{dx}{1+\sin x}\\ I=\frac{\pi }{2}{\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\left(\frac{1-\sin x}{{\cos }^{2}x}\right)dx\\ I=\frac{\pi }{2}{\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\left({\sec }^{2}x-\tan x\sec x\right)dx\\ =I-\frac{\pi }{2}{\left[\tan x-\sec x\right]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\\ =\frac{\pi }{2}\left[\left(-1+\sqrt{2}\right)-\left(1-\sqrt{2}\right)\right]\\ I=\pi \left[\left(\sqrt{2}-1\right)\right]\\ OptionBiscorrectanswer.\end{array}$