1. ## Interesting Integrals. #1

Evaluate:

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

2. Put $x\mapsto \pi - x$ to get
$\int_0^{\pi} \frac{x \sin x }{1+\cos^2 x} dx = \int_0^{\pi} \frac{(\pi-x) \sin x }{1+\cos^2 x} dx = -\int_0^{\pi} \frac{x \sin x }{1+\cos^2 x} dx + \pi \int_0^{\pi} \frac{\sin x }{1+\cos^2 x} dx$
And thus
$\int_0^{\pi} \frac{x \sin x }{1+\cos^2 x} dx = \frac{\pi}{2} \int_0^{\pi} \frac{\sin x }{1+\cos^2 x} dx$
The latter integral can easily be evaluated as $\frac{\pi}{2}$ by letting $x \mapsto \cos x$. Thus your integral is:
$\int_0^{\pi} \frac{x \sin x }{1+\cos^2 x} dx = \frac{\pi^2}{4}$

3. Integrating by parts You obtain first...

$\int_{0}^{\pi} \frac{x\cdot \sin x}{1+\cos^{2} x}\cdot dx = |-x\cdot \tan^{-1} \cos x|_{0}^{\pi} + \int_{0}^{\pi} \tan^{-1} \cos x\cdot dx$ (1)

Now You perform in the second integral the substitution $\xi= x - \frac{\pi}{2}$ so that it becomes...

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan^{-1} \sin \xi\cdot d\xi$ (2)

Now the integrand function in (2) is an odd function, so that the integral (2) vanishes and is...

$\int_{0}^{\pi} \frac{x\cdot \sin x}{1+\cos^{2} x}\cdot dx = |-x\cdot \tan^{-1} \cos x|_{0}^{\pi} = \frac{\pi^{2}}{4}$ (3)

Kind regards

$\chi$ $\sigma$

4. write $\int_{0}^{\pi }{\arctan (\cos x)\,dx}=\int_{0}^{\pi }{\int_{0}^{1}{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dt}\,dx}=\int_{0}^{1}{\int_{0}^{\pi }{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dx}\,dt},$ and in the last integral substitute $x\mapsto \pi-x,$ you'll see it vanishes as well.

5. ## i donot know double integrals

Originally Posted by Krizalid
write $\int_{0}^{\pi }{\arctan (\cos x)\,dx}=\int_{0}^{\pi }{\int_{0}^{1}{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dt}\,dx}=\int_{0}^{1}{\int_{0}^{\pi }{\frac{\cos x}{1+t^{2}\cos ^{2}x}\,dx}\,dt},$ and in the last integral substitute $x\mapsto \pi-x,$ you'll see it vanishes as well.