Question

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

Answer 1

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

$={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{xdx}{1+\sin x}$ ......$\left(1\right)$

Replace $x\to \frac{\pi }{4}+\frac{3\pi }{4}-x=\pi -x$

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

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

Adding $\left(1\right)$ and $\left(2\right)$ we get

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

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

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

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

Let $t=x-\frac{\pi }{2}$

$2I=\pi {\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\frac{dt}{1+\cos t}$

$\Rightarrow 2I=2\pi {\int }_0^{\frac{\pi }{4}}\frac{dt}{1+\cos t}$

$\Rightarrow 2I=2\pi {\int }_0^{\frac{\pi }{4}}{\sec }^2\frac{t}{2}dt$

$\Rightarrow 2I=2\pi {\left[\tan \frac{t}{2}\right]}_0^{\frac{\pi }{4}}$

$\Rightarrow 2I=2\pi \tan \frac{\pi }{8}$

$\Rightarrow 2I=2\pi (\sqrt{2}-1)$

$\therefore I=\pi (\sqrt{2}-1)$