Incomprehension in a variable change of an integral to compute the Bâle's Problem

Question

In this topic they can read this answer, and I don't understand in the proof (at page 32) how they changed the bond of the integral.

The integral is : $$4\int_{\frac{1}{2}}^1 \arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)\cdot \frac{\mathrm{d}u}{\sqrt{1-u^2}}$$

Then the variable change is $u=\cos(2\theta)$ then $\mathrm{d}u=-2\sin(2\theta)\mathrm{d}\theta$

And I don't understand why the following integral is $$4\int_{0}^{\frac{\pi}{6}} ...$$

For me when $u$ goes from $1/2$ to $1$, $\theta$ doesn't go from $0$ to $\pi/6$...

Miss-I something? I'm ashamed to ask this question but after some days of think, it makes me a bit sad... :(

Thank you in advance.

I don't understand what you're saying. Use what I wrote, then reverse the order of integration and the minus sign disappears. — zhw., Sep 21 '15 at 19:15
If $u=1$ using the var. change, they have $1=\cos(2*\theta)$ then $\theta=\pi/6$??????? — ParaH2, Sep 21 '15 at 19:30

score 0 · Answer 1 · answered Sep 21 '15 at 20:12

0

$$\int_{\frac{1}{2}}^1 f(u)\,du = \int_{\pi/6}^0 f(\cos 2t)(-2\sin 2t)\, dt = \int_{0}^{\pi/6} f(\cos 2t)(2\sin 2t)\, dt.$$

answered Sep 21 '15 at 20:12

zhw.

I know that just what i don't understand is the value $0$ and $\pi/6$ to integrate. :/ – ParaH2 Sep 21 '15 at 20:34

1 Answers1