Question

Integrate:

${\int }_0^{\pi }\frac{xdx}{1+\sin x}$ ?

Answer 1

Let the given integral be,

$I={\int }_0^{\pi }\frac{xdx}{1+\sin x}⟶\left(1\right)$

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

$\therefore I={\int }_0^{\pi }\frac{\pi -x}{1+\sin x}dx⟶\left(2\right)$

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

$I+I={\int }_0^{\pi }\frac{x}{1+\sin x}dx+{\int }_0^{\pi }\frac{\pi -x}{1+\sin x}dx\Rightarrow 2I={\int }_0^{\pi }\frac{x+\pi -x}{1+\sin x}dx\Rightarrow 2I={\int }_0^{\pi }\frac{\pi }{1+\sin x}dx\Rightarrow I=\frac{\pi }{2}{\int }_0^{\pi }\frac{1}{1+\sin x}dx\therefore I=\frac{\pi }{2}{\int }_0^{\pi }\frac{1}{1+\sin x}dx=\frac{\pi }{2}{\int }_0^{\pi }(\frac{1}{1+\sin x}\times \frac{1-\sin x}{1-\sin x})dx=\frac{\pi }{2}{\int }_0^{\pi }\left(\frac{1-\sin x}{1-{\sin }^{2}x}\right)dx=\frac{\pi }{2}{\int }_0^{\pi }\left(\frac{1-\sin x}{{\cos }^{2}x}\right)dx=\frac{\pi }{2}{\int }_0^{\pi }[\frac{1}{{\cos }^{2}x}-\frac{\sin x}{{\cos }^{2}x}]dx=\frac{\pi }{2}{\int }_0^{\pi }[{\sec }^{2}x-\frac{\sin x}{\cos x\cos x}]dx=\frac{\pi }{2}{\int }_0^{\pi }[{\sec }^{2}x-\tan x\sec x]dx=\frac{\pi }{2}[{\int }_0^{\pi }{\sec }^{2}xdx-{\int }_0^{\pi }\tan x\sec xdx]=\frac{\pi }{2}[{\left[\tan x\right]}_0^{\pi }-{\left[\sec x\right]}_0^{\pi }]=\frac{\pi }{2}[[\tan \pi -\tan 0]-[\sec \pi -\sec 0]]=\frac{\pi }{2}[[0-0]-[-1-1]]=\frac{\pi }{2}[0-(-2\left)\right]=\frac{\pi }{2}\times 2=\pi \therefore {\int }_0^{\pi }\frac{xdx}{1+\sin x}=\pi .$