0

Give the expression of the integral $$\int_0^\pi\frac{dx}{1+\alpha^2\sin^2(x)}$$ where $ \alpha \in (0,+\infty)$.

I tried the substitution $ t=\tan(x)$ but both bounds become zero. I used $t=\tan(\frac x2) $ but it became complicate, i think there is an easier way.

Thank you in advance.

hamam_Abdallah
  • 62,951
  • 2
    When doing a substitution, don't forget to check it is one-to-one (and do the needed adjustments if it is not). In your case, you can notice that $\int_0^\pi=2\int_0^{\pi/2}$ (and then do the first substitution). – metamorphy May 26 '20 at 13:48
  • 4
    In this case, it is not being one-to-one that is the problem, rather it is the singularity of $\tan(x)$ at $x = \pi/2$. – Robert Israel May 26 '20 at 13:52

2 Answers2

1

Your first substitution of $t=\tan{x}$ is correct, but as @metamorphy said, you must adjust the bounds to because it is not one to one with the $\tan{x}$ substitution (and this works because of symmetry with $\sin^2{x}$ from $0$ to ${\pi}$ is twice of $\sin^2{x}$ from $0$ to $\frac{\pi}{2}$.

$\frac{dt}{t^2+1}=dx$ and $\sin^2{x}=\frac{t^2}{t^2+1}$:

$I=2\int_0^{\infty} \frac{\frac{dt}{t^2+1}}{1+\frac{a^2t^2}{t^2+1}}=2\int_0^{\infty} \frac{dt}{\left(a^2+1\right)t^2+1}=\frac{2}{\sqrt{a^2+1}}\arctan{\left(t\sqrt{a^2+1}\right)} \big\rvert_0^{\infty}=\frac{2}{\sqrt{a^2+1}} \cdot \frac{\pi}{2}=\boxed{\frac{\pi}{\sqrt{a^2+1}}}$

Hope this helps.

Ty.
  • 5,434
1

\begin{align*}\int_0^\pi\frac{dx}{1+\alpha^2\sin^2x}&=\int_0^\pi\frac{dx}{\cos^2x+(\alpha^2+1)\sin^2x} \\ &=\int_0^{\frac{\pi}{2}}\frac{2\sec^2(x)}{1+(\alpha^2+1)\tan^2x}dx \\ & = \frac{2}{\sqrt{\alpha^2+1}}\int_0^{\frac{\pi}{2}}\frac{d\:(\sqrt{\alpha^2+1}\tan x)}{1+(\alpha^2+1)\tan^2x} \\ & = \frac{2}{\sqrt{\alpha^2+1}} \arctan(\alpha\tan x)\large\rvert_{\small0}^{\small\frac{\pi}{2}} \\ & = \frac{\pi}{\sqrt{\alpha^2+1}} \end{align*}

Alan
  • 2,025