Kiana Dodson

2022-06-26

Jump discontinuity in antiderivative
I am dealing with integral $I={\int }_{0}^{\pi }dx\frac{{\mathrm{sin}}^{2}x}{{a}^{2}+{\mathrm{sin}}^{2}x}.$
I know the answer is $I=\pi \left(1-\frac{a}{\sqrt{1+{a}^{2}}}\right).$.
However I am having some uncertainty getting there. I know that before plugging in the integration limits the solution looks like $I={\left[x-\frac{a{\mathrm{tan}}^{-1}\left(\frac{\sqrt{1+{a}^{2}}}{a}\mathrm{tan}x\right)}{\sqrt{1+{a}^{2}}}\right]}_{x=0}^{x=\pi }$.
but at first sight this would yield merely $\pi$. I can see that the right step to recover the correct solution is to take ${\mathrm{tan}}^{-1}\left(\frac{\sqrt{1+{a}^{2}}}{a}\mathrm{tan}x\right){|}_{x=\pi }=\pi$ however how do I justify this (besides that it yields the correct answer)?

Blaine Foster

Step 1
First notice that the integral exists for all $a\in \mathbb{R}$, and from symmetry
$I\left(a\right)=2{\int }_{0}^{\pi /2}\frac{{\mathrm{sin}}^{2}x}{{a}^{2}+{\mathrm{sin}}^{2}x}dx\phantom{\rule{0ex}{0ex}}=\pi -2{a}^{2}{\int }_{0}^{\pi /2}\frac{dx}{{a}^{2}+{\mathrm{sin}}^{2}x}$
Then we have that ${\mathrm{sin}}^{2}x=\frac{1}{2}\left(1-\mathrm{cos}2x\right)$ so
$I\left(a\right)=\pi -4{a}^{2}{\int }_{0}^{\pi /2}\frac{dx}{2{a}^{2}+1-\mathrm{cos}2x}\phantom{\rule{0ex}{0ex}}=\pi +2{a}^{2}{\int }_{0}^{\pi }\frac{dx}{\mathrm{cos}x-2{a}^{2}-1}$
Step 2
Then from the comments here, we have that
${\int }_{0}^{\pi }\frac{dx}{b+a\mathrm{cos}x}=\frac{\pi }{\sqrt{{b}^{2}-{a}^{2}}}$
(I can show you a proof if you'd like). So of course
$I\left(a\right)=\pi +\frac{2{a}^{2}\pi }{\sqrt{\left(2{a}^{2}+1{\right)}^{2}-1}}=\pi +\frac{a\pi }{\sqrt{{a}^{2}+1}}$
And from the definition of I(a), we have that $I\left(a\right)=I\left(-a\right)$ so your desired result:
$\frac{1}{\pi }I\left(a\right)=1-\frac{a}{\sqrt{{a}^{2}+1}}$

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