How do you solve the integral int_{0}^{pi/2} dx/(2 + cosx) ?

Question

he66666 · Answer

Sorry I'll type again

he66666 · Answer

$$\int\limits_{0}^{\pi/2} dx/(2 + cosx) $$

he66666 · Answer

the answer is $$\pi/(3\sqrt{3})$$

he66666 · Answer

I tried partial integration by u = cosx + 2 and du = -sinxdx but it didn't work out..

anonymous · Answer

there's a tricky substitution that pretty much nobody who hasn't seen the integral before would see, but your substitution will do just fine.
after you do that substitution (i'm ignoring the bounds for now):
int -(sin(x)*u)^-1, but sin(x) = sqrt(1 -cos^2(x)) = sqrt(1 - (u -2)^2).
which completely transforms the integral, try giving that a go

he66666 · Answer

I converted the expression to the way you explained but I only got $$\int\limits_{}^{}1/(u \sqrt{1-(u-2)^2)}du$$ and I don't know what to do now

anonymous · Answer

just to tidy things up you could do another substitution z = u - 2 -> dz = du, u = z + 2
so int((z+2)sqrt(1 - z^2))^-1 dz, which you could try to do by parts, i'll check that though

he66666 · Answer

Weirdly, I got pi as the answer...

anonymous · Answer

Gaah, openstudy is pissing me off. I will try to write in latex, hang on.

anonymous · Answer

Again, I am unsure if this will be comprehensible. Sorry for not trying to clean this up :-) But, yeah, I think it's solvable now.

he66666 · Answer

Thank you!