# Compute the definite integral

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

I tried letting $u=\tan(\frac{x}{2})$ but could not make it work. A few other trig substitutions failed as well. I noticed the integrand is odd but could not make use of that fact. Any ideas?

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Note that the denominator is symmetric with respect to $\frac{\pi}{2}$. So we have

$$\int_0^\pi \frac{x}{1+\cos^2 x}\,dx = \frac{\pi}{2}\int_0^\pi \frac{1}{1+\cos^2 x}\,dx + \int_0^\pi \frac{x-\frac{\pi}{2}}{1+\cos^2 x}\,dx,$$

and the integrand of the second integral is "odd with respect to $\frac{\pi}{2}$", so

$$\int_0^\pi \frac{x}{1+\cos^2 x}\,dx = \frac{\pi}{2}\int_0^\pi \frac{1}{1+\cos^2 x}\,dx.$$

We can evaluate that integral using the residue theorem similar to this, and obtain

$$\int_0^\pi \frac{x}{1+\cos^2 x}\,dx = \frac{\pi^2}{2\sqrt{2}}.$$

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Should be $\frac{\pi^2}{2 \sqrt{2}}$? –  mtiano Jan 15 at 16:01
Sure. Typo. I had it right when checking that I didn't miscompute, Prelude> pi^2/(2*sqrt 2) 3.4894320998194392, but mistyped nevertheless :( –  Daniel Fischer Jan 15 at 16:03

Break up into $A$, the integral from $0$ to $\pi/2$, plus $B$, the integral from $\pi/2$ to $\pi$, For the second integral, make the change of variable $x=\pi -u$. We get $$B=\int_{x=\pi/2}^{\pi} \frac{x}{1+\cos^2 x}\,dx=\int_{\pi/2}^{0} -\frac{\pi-u}{1+\cos^2 u}\,du.$$ With sign changes, and replacement of the letter $u$ by the letter $x$, we get that $$B=\int_{0}^{\pi/2} \frac{\pi-x}{1+\cos^2 x}\,dx.$$ Add $A$. We get that $$A+B=\int_0^{\pi/2}\frac{\pi}{1+\cos^2 x}\,dx.$$ Now we are at a moderately standard integral.

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use this trigonometry identity

$\cos^2(x) = \frac{1}{2}+\frac{1}{2}\cos(2x)$

you will get

$\dfrac{x}{3/2+cos(2x)/2}$

now

this is same as

$\dfrac{2x}{3+\cos(2x)}$

please now substitute $y=2x$

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no no,i have changed –  dato datuashvili Jan 15 at 15:55
it should be $1/2$ multiply by cos(2x) –  dato datuashvili Jan 15 at 15:56
but can it computed by elementary fucntion? –  dato datuashvili Jan 15 at 16:00

As $\displaystyle\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$

$I=\displaystyle\int_0^\pi\frac x{1+\cos^2x}dx=\int_0^\pi\frac{\pi-x}{1+\cos^2x}dx$ as $\cos(\pi-x)=-\cos x$

$\displaystyle\implies I+I=\pi\int_0^\pi\frac1{1+\cos^2x}dx$

Now $\displaystyle \int_0^{2a}f(x)dx= \begin{cases} 2\int_0^af(x)dx &\mbox{if } f(2a-x)=f(x) \\ 0 & \mbox{if } f(2a-x)=-f(x) \end{cases}$

Here $\displaystyle a=\frac\pi2\implies \int_0^\pi\frac1{1+\cos^2x}dx=2\int_0^{\frac\pi2}\frac1{1+\cos^2x}dx$

Now $\displaystyle\int\frac1{1+\cos^2x}dx=\int\frac{\sec^2x}{2+\tan^2x}dx$

Set $\tan x=u$

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