1. ## Integral

how I can integrate this

$\displaystyle \int_{0}^{\pi} \frac{x\sin x }{1 + \cos ^2 x } dx$

any clues hints

2. Using $\displaystyle \ \int_{a}^{b}f(x)\;{dx}=\int_{a}^{b}f(a+b-x)\;{dx}\$, we have:

\begin{aligned}\displaystyle I & = \int_{0}^{\pi}\frac{x\sin{x}}{1+\cos^2{x}}\;{dx} = \int_{0}^{\pi}\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}}\;{dx} \\ & = \int_{0}^{\pi}\frac{\pi\sin(\pi-x)}{1+\cos^2(\pi-x)}}\;{dx}-\int_{0}^{\pi}\frac{x\sin(\pi-x)}{1+\cos^2(\pi-x)}}\;{dx} \\ & = \pi\int_{0}^{\pi}\frac{\sin{x}}{1+\cos^2{x}}}\;{dx }-\int_{0}^{\pi}\frac{x\sin{x}}{1+\cos^2{x}}}\;{dx} \\ & = \pi\int_{0}^{\pi}\frac{\sin{x}}{1+\cos^2{x}}}\;{dx }-I .\end{aligned}

Rearranging this, we get:

$\displaystyle 2I = \pi\int_{0}^{\pi}\frac{\sin(x)}{1+\cos^2(x)}\;{dx} .$

Therefore, we have:

\displaystyle \begin{aligned} I & = \frac{\pi}{2}\int_{0}^{\pi}\frac{\sin{x}}{1+\cos^2 {x}}\;{dx} = -\frac{\pi}{2}\int_{0}^{\pi}\frac{(\cos{x})'}{1+\co s^2{x}}\;{dx} \\ & = -\frac{\pi}{2}\tan^{-1}\left(\cos{x}\right)\bigg|_{0}^{\pi} = -\frac{\pi}{2}\left(-\frac{\pi}{2}\right) = \frac{\pi^2}{4}.\end{aligned}

3. Tanks very much I appreciate that

$\displaystyle \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$

or the proof

4. Originally Posted by Amer
$\displaystyle \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ or the proof

Using the substitution

$t=a+b-x$

you'll obtain

$\displaystyle \int_{a}^{b} f(a+b-x) dx =\ldots=\displaystyle \int_{a}^{b} f(t) dt=\displaystyle \int_{a}^{b} f(x) dx$

Fernando Revilla

5. Originally Posted by FernandoRevilla
Using the substitution

$t=a+b-x$
thanks very much

6. integrate by parts and get $\displaystyle\frac{{{\pi }^{2}}}{4}+\int_{0}^{\pi }{\arctan (\cos x)\,dx},$ now the last integral is zero by putting $t=x-\dfrac\pi2$ so the latter one equals $\displaystyle-\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\arctan (\sin x)\,dx}=0,$ since the function is odd.

or even easier, put $t=\pi-x$ and see that the last integral equals to its negative, hence it must be zero.

7. In general, $\displaystyle \int^{\pi}_{0} x f(\sin x) \ dx = \frac{\pi}{2} \int^{\pi}_{0} f(\sin x) \ dx$.

For this problem $\displaystyle f(\sin x) = \frac{\sin x}{2- \sin^{2} x}$

8. Originally Posted by Krizalid
integrate by parts and get $\displaystyle\frac{{{\pi }^{2}}}{4}+\int_{0}^{\pi }{\arctan (\cos x)\,dx},$ now the last integral is zero by putting $t=x-\dfrac\pi2$ so the latter one equals $\displaystyle-\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\arctan (\sin x)\,dx}=0,$ since the function is odd.

or even easier, put $t=\pi-x$ and see that the last integral equals to its negative, hence it must be zero.
Thanks thats great