Question

Evaluate:∫(0->π ) (x sinx)/(1+cos^2x) dx

Answer 1

It can be solved in two ways, I've mentioned both. Ask if you have any doubt at any step.

Answer 2

The below mentioned answer is not correct. Because the denominator is $1+\cos^2 x$ not $1+\cos 2x$. Immediately delete that answer.

Use the property of the definite integral,

$\int\limits^a_0 {f(x)} \, dx =\int\limits^a_0 {f(a-x)} \, dx$.

Here we have to find,

$\int\limits^{\pi}_0 {\frac{x\sin x}{1+\cos^2 x}} \, dx$.

Now using the property of the integrals,

$I=\int\limits^{\pi}_0 {\frac{x\sin x}{1+\cos^2 x}} \, dx .....................(1)\\ I=\int\limits^{\pi}_0 {\frac{(\pi-x)\sin (\pi-x)}{1+\cos^2 (\pi-x)}} \, dx \\ I=\int\limits^{\pi}_0 {\frac{(\pi-x)\sin x}{1+\cos^2 x}} \, dx ........................(2)$

Adding equations (1) and (2),

$2I=\int\limits^{\pi}_0 {\frac{\pi\sin x}{1+\cos^2 x}} \,dx \\ 2I=\pi\int\limits^{\pi}_0 {\frac{\sin x}{1+\cos^2 x}} \,dx\\ 2I=-\pi [\tan^{-1}(\cos x)]^{\pi}_0\\ 2I=-\pi [\tan^{-1}(-1)-\tan^{-1}(0)]\\ 2I=-\pi *(\frac{\pi}{4})\\ 2I=\frac{\pi^2}{4}\\ I=\frac{\pi^2}{8}$