Integrate xsin(x) ∕ (1+cos^2(x)) dx from 0 to pi?

Question

If it helps, here's my attempt at it,

xsin(x) ∕ (1+cos^2(x))

=xsin(x) ∕ (2−sin^2(x))

=xsin(x) ∕ [(sqrt(2)−sin(x))(sqrt(2)+sin(x))]

Now using partial fractions, get

=x ∕ (2sqrt(2)−2sin(x))

−x ∕ (2sqrt(2)+2sin(x))

but i'm not sure if this gets us anywhere.

Thanks to everyone who thought about this problem.

Answer 1

[Edited]

We can use integration by parts to simplify the integral:

f' = sin x / (1 + cos^2 x)

g = x

g' = 1

For finding f, use

u = cos x

du = -sin x dx

dx sin(x) / (1 + cos^2(x)) = -du/(1+u^2)

=> f = -arctan cos x

Our integral then becomes

-x arctan cos x + int(arctan cos x, x = 0..pi);

The second part is zero, which we can find easily by shifting x by -pi/2 which makes it an integral of an odd function in symmetric bounds (or we can sketch an image ;-))

Plugging in zero and pi, we obtain the result

-0 * arctan cos 0 + pi * arctan cos pi = pi*arctan(-1) = pi^2/4.

Hope this helps ;-)