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I am trying to solve the following definite integral:

$$\int_0^\pi \frac{x\sin(x)}{1+\cos^2(x)}dx$$

However, it appears that an elementary antiderivative does not exist for the function. Is there any way to solve the definite integral?

Em.
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James
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    $\int_0^\pi f(x)dx=\int_0^\pi f(\pi-x)dx$. You wouldn't need to find the actual anti derivative in this manner. My suggestion is laid out in this similar problem: https://math.stackexchange.com/questions/1100566/how-to-solve-int-0-pi-fracx-sin-x1-sin2-xdx?rq=1 – imranfat Jun 29 '17 at 04:31
  • this was asked billions of times before :( – tired Jun 29 '17 at 07:00

1 Answers1

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Let the integral be $I $ using $\int _0^{\pi}f (x)=\int_0^{\pi}f (\pi-x)$ .We have $$\text{I}=\int_0^{\pi} \frac {\pi\sin (x)}{1+\cos^2 (x)}-\text {I}$$. Therefore $2\text {I}=\int_0^{\pi} \frac {\pi\sin (x)}{1+\cos^2 (x)} =\frac {\pi^2}{2}$ thus $I=\frac {\pi^2}{4} $.

Archis Welankar
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