Is $\int_{0}^{\pi}{x\over a+b\tan^2x}\mathrm dx=\int_{0}^{\pi}{x\over a+b\cot^2x}\mathrm dx?$

Question

Consider these two integrals:

$$\int_{0}^{\pi}{x\over a+b\tan^2x}\mathrm dx=\int_{0}^{\pi}{x\over a+b\cot^2x}\mathrm dx$$ where $(a,b)$ are real numbers

Are they equal, because it is trivial? I can't see it.

Can anyone demonstrate how they are equal?

@labbhattacharjee, I think you need a - in front of your 2nd integral, — user5389726598465, Mar 29 '17 at 05:28
I don't think the equality is true, although I can't prove it. — jonsno, Mar 29 '17 at 05:37
$\tan(x) = -\cot(x+\pi/2)$ and both are $\pi$-periodic @user135711 right at its zero $\cot'(z_0)=-1$ — reuns, Mar 29 '17 at 05:43
@user135711, See http://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx/439856#439856 — lab bhattacharjee, Mar 29 '17 at 05:46
I'm just distracting from the question. @lab ahh, now I see. very correct — user5389726598465, Mar 29 '17 at 05:51

score 2 · Accepted Answer · answered Mar 29 '17 at 13:21

By splitting the integration range in halves the first integral equals: $$ 2\int_{0}^{\pi/2}\frac{dx}{a+b\tan^2(x)} = 2\int_{0}^{+\infty}\frac{dt}{(a+bt^2)(1+t^2)}$$ and the same applies to the second integral. Both equal $\int_{\mathbb{R}}\frac{dt}{(a+bt^2)(1+t^2)}$.

Is $\int_{0}^{\pi}{x\over a+b\tan^2x}\mathrm dx=\int_{0}^{\pi}{x\over a+b\cot^2x}\mathrm dx?$

1 Answers1