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 $I=\pi (1-\frac{a}{\sqrt{1+{a}^{2}}}).$.

However I am having some uncertainty getting there. I know that before plugging in the integration limits the solution looks like $I={[x-\frac{a{\mathrm{tan}}^{-1}(\frac{\sqrt{1+{a}^{2}}}{a}\mathrm{tan}x)}{\sqrt{1+{a}^{2}}}]}_{x=0}^{x=\pi}$.

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}(\frac{\sqrt{1+{a}^{2}}}{a}\mathrm{tan}x){{\textstyle |}}_{x=\pi}=\pi $ however how do I justify this (besides that it yields the correct answer)?

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 $I=\pi (1-\frac{a}{\sqrt{1+{a}^{2}}}).$.

However I am having some uncertainty getting there. I know that before plugging in the integration limits the solution looks like $I={[x-\frac{a{\mathrm{tan}}^{-1}(\frac{\sqrt{1+{a}^{2}}}{a}\mathrm{tan}x)}{\sqrt{1+{a}^{2}}}]}_{x=0}^{x=\pi}$.

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}(\frac{\sqrt{1+{a}^{2}}}{a}\mathrm{tan}x){{\textstyle |}}_{x=\pi}=\pi $ however how do I justify this (besides that it yields the correct answer)?

Blaine Foster

Beginner2022-06-27Added 33 answers

Step 1

First notice that the integral exists for all $a\in \mathbb{R}$, and from symmetry

$I(a)=2{\int}_{0}^{\pi /2}\frac{{\mathrm{sin}}^{2}x}{{a}^{2}+{\mathrm{sin}}^{2}x}dx\phantom{\rule{0ex}{0ex}}=\pi -2{a}^{2}{\int}_{0}^{\pi /2}\frac{dx}{{a}^{2}+{\mathrm{sin}}^{2}x}$

Then we have that ${\mathrm{sin}}^{2}x=\frac{1}{2}(1-\mathrm{cos}2x)$ so

$I(a)=\pi -4{a}^{2}{\int}_{0}^{\pi /2}\frac{dx}{2{a}^{2}+1-\mathrm{cos}2x}\phantom{\rule{0ex}{0ex}}=\pi +2{a}^{2}{\int}_{0}^{\pi}\frac{dx}{\mathrm{cos}x-2{a}^{2}-1}$

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

$I(a)=\pi +\frac{2{a}^{2}\pi}{\sqrt{(2{a}^{2}+1{)}^{2}-1}}=\pi +\frac{a\pi}{\sqrt{{a}^{2}+1}}$

And from the definition of I(a), we have that $I(a)=I(-a)$ so your desired result:

$\frac{1}{\pi}I(a)=1-\frac{a}{\sqrt{{a}^{2}+1}}$

First notice that the integral exists for all $a\in \mathbb{R}$, and from symmetry

$I(a)=2{\int}_{0}^{\pi /2}\frac{{\mathrm{sin}}^{2}x}{{a}^{2}+{\mathrm{sin}}^{2}x}dx\phantom{\rule{0ex}{0ex}}=\pi -2{a}^{2}{\int}_{0}^{\pi /2}\frac{dx}{{a}^{2}+{\mathrm{sin}}^{2}x}$

Then we have that ${\mathrm{sin}}^{2}x=\frac{1}{2}(1-\mathrm{cos}2x)$ so

$I(a)=\pi -4{a}^{2}{\int}_{0}^{\pi /2}\frac{dx}{2{a}^{2}+1-\mathrm{cos}2x}\phantom{\rule{0ex}{0ex}}=\pi +2{a}^{2}{\int}_{0}^{\pi}\frac{dx}{\mathrm{cos}x-2{a}^{2}-1}$

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

$I(a)=\pi +\frac{2{a}^{2}\pi}{\sqrt{(2{a}^{2}+1{)}^{2}-1}}=\pi +\frac{a\pi}{\sqrt{{a}^{2}+1}}$

And from the definition of I(a), we have that $I(a)=I(-a)$ so your desired result:

$\frac{1}{\pi}I(a)=1-\frac{a}{\sqrt{{a}^{2}+1}}$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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