Question

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

• A. $\frac{{\pi }^2}{4}-(\sqrt{2}-1)\pi$
• B. $\frac{{\pi }^2}{4}-(\sqrt{2}+1)\pi$
• C. $\frac{{\pi }^2}{2}-(\sqrt{2}-1)\pi$
• D. $\frac{{\pi }^2}{2}-(\sqrt{2}+1)\pi$

Answer 1

The correct option is A $\frac{{\pi }^2}{4}-(\sqrt{2}-1)\pi$

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

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

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

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

$\Rightarrow I=\frac{\pi }{2}{\left[x\right]}_{\pi /4}^{3\pi /4}-\frac{\pi }{2}{\int }_{\pi /4}^{3\pi /4}\frac{1-\sin x}{{\cos }^{2}x}dx$

$=\frac{{\pi }^2}{4}-\frac{\pi }{2}{[\tan x-\sec x]}_{\pi /4}^{3\pi /4}=\frac{{\pi }^2}{4}-\frac{\pi }{2}[(-1-1)-(-\sqrt{2}-\sqrt{2})]$

$\therefore I=\frac{{\pi }^2}{4}+\frac{\pi }{2}(2-2\sqrt{2})=\frac{{\pi }^2}{4}-(\sqrt{2}-1)\pi$