Interesting integral $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}$

Question

Latest Edit

Inspired by @J.G., I find a formula in general,

$$ \begin{aligned} \int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{n}} &=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{n}} \\ &=\left.\frac{2(-1)^{n}}{(n-1) !} \frac{\partial^{n}}{\partial a^{n}}\left(\int_{0}^{\pi} \frac{d x}{a-\cos x} \right)\right|_{a=3} \\ &=\left.\frac{2(-1)^{n} \pi}{(n-1) !} \frac{\partial^{n}}{\partial a^{n}}\left(\frac{1}{\sqrt{a^{2}-1}}\right)\right|_{a=3} \end{aligned} $$

Multiplying both the numerator and denominator $\sec^4x$ yields

$\displaystyle I=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{4} x}{\left(\sec ^{2} x+\tan ^{2} x\right)^{2}} d x \tag*{} $ Letting $ t=\tan x$ gives $\displaystyle \begin{aligned}I&= \int_{0}^{\infty} \frac{1+t^{2}}{\left(1+2 t^{2}\right)^{2}} d t\\&=\int_{0}^{\infty} \frac{1+\frac{1}{t^{2}}}{\left(2 t+\frac{1}{t}\right)^{2}} d t\\&=\int_{0}^{\infty} \frac{\frac{3}{4}\left(2+\frac{1}{t^{2}}\right)-\frac{1}{4}\left(2-\frac{1}{t^{2}}\right)}{\left(2 t+\frac{1}{t}\right)^{2}} d t\\&=\frac{3}{4} \int_{0}^{\infty} \frac{d\left(2 t-\frac{1}{t}\right)}{\left(2 t-\frac{1}{t}\right)^{2}+8}-\frac{1}{4} \int_{0}^{\infty} \frac{d\left(2 t+\frac{1}{t}\right)}{\left(2 t+\frac{1}{t}\right)^{2}}\\&=\left[\frac{3}{8 \sqrt{3}} \tan ^{-1}\left(\frac{2 t-\frac{1}{t}}{2 \sqrt{2}}\right)+\frac{1}{4\left(2 t+\frac{1}{t}\right)}\right]_{0}^{\infty}\\&=\frac{3 \pi}{8 \sqrt{2}}\end{aligned}\tag*{} $

Is there any method other than tangent half-angle substitution?

Extending the integral to $(-\pi,\pi)$ and using the residue theorem should work. — Diger, Aug 08 '22 at 09:42
@Diger Do you mean using $I$ as a quarter-period rather than the half-period it constitutes in my answer? If so, you're right, it will work. Unfortunately, it's much messier, because you need a second-order pole of a quartic, which is squared in the resulting integrand's denominator. — J.G., Aug 08 '22 at 09:49
Yes, using the quarter period. I hadn't seen your answer when I posted; I like it better to transform $2\sin^2(x)=1-\cos(2x)$ first though. — Diger, Aug 08 '22 at 10:07

Quanto · Answer 1 · 2022-08-10T15:07:10.577

7

\begin{align} \int_{0}^{\frac{\pi}{2}} \frac{dx}{(1+\sin ^{2} x)^{2}} = &\int_{0}^{\frac{\pi}{2}} \frac1{4\tan^3x} \ d\left(\frac{\tan^2x}{1+\csc^2x}\right)\\ \overset{ibp}=&\ \frac34 \int_{0}^{\frac{\pi}{2}} \frac{d(\tan x)}{1+2\tan^2x} =\frac{3\pi}{8\sqrt2} \end{align}

edited Aug 10 '22 at 15:07

answered Aug 08 '22 at 15:28

Quanto

97,352

How did you spot the $d(\tan^2/(1+\csc^2x))$ term? Incredible – FShrike Aug 10 '22 at 15:11
@FShrike - not obvious, granted. I’m a fun of ibp; knowing some obscure dv’s from experience – Quanto Aug 10 '22 at 15:15
1

You could probably also guess a substitution from Bioche's rules. – J.G. Aug 10 '22 at 18:32
1

Your first line is inconceivable. – Lai Aug 12 '22 at 00:57
@Quanto Out of curiosity, what other obscure dv's do you know from experience? – Frank W Aug 13 '22 at 02:24
@FrankW - here we go \begin{align} \int_0^{\pi/4} \cos^2x\sin^2 x \ dx =& -\frac1{16}\int_0^{\pi/4} \tan 2x\ d(\cos^2 2x)= \frac18 \int_0^{\pi/4}dx=\frac{\pi}{32} \end{align} – Quanto Aug 13 '22 at 03:17

score 5 · Accepted Answer · answered Aug 08 '22 at 09:33

The obvious alternative is the residue theorem. With $y=2x$ your integral is $$\int_0^\pi\frac{2dy}{(3-\cos y)^2}=\int_0^{2\pi}\frac{dy}{(3-\cos y)^2}\stackrel{z=e^{iy}}{=}\oint_{|z|=1}\frac{-4izdz}{(z^2-6z+1)^2}.$$There is one enclosed pole, the second-order $3-2\sqrt{2}$, so the integral is$$8\pi\lim_{z\to3-2\sqrt{2}}\frac{d}{dz}\frac{z}{(z-3-2\sqrt{2})^2}=8\pi\lim_{z\to3-2\sqrt{2}}\frac{z+3+2\sqrt{2}}{(3+2\sqrt{2}-z)^3}=\frac{3\pi}{8\sqrt{2}}.$$

Thank you for your interesting and short solution! – Lai Aug 08 '22 at 09:44 — Lai, Aug 08 '22 at 09:44

Lai · Answer 3 · 2022-08-12T13:19:08.390

Start as J.G., then using differentiation yields $$ \begin{aligned} \int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\frac{1-\cos 2 x}{2}\right)^{2}} &=4 \int_{0}^{\frac{\pi}{2}} \frac{d x}{(3-\cos 2 x)^{2}} \\ &=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{2}} \\ &=-\left.2 \frac{\partial}{\partial a} \int_{0}^{\pi} \frac{d x}{a-\cos x}\right|_{a=3} \\ &=-\left.2 \frac{\partial}{\partial a}\left(\frac{\pi}{\sqrt{a^{2}-1}}\right)\right|_{a=3} \\ &=\frac{3 \pi}{8 \sqrt{2}} \end{aligned} $$

where the last second line comes from my post$\int_{0}^{\pi} \frac{d \theta}{a-b \cos \theta} =\frac{\pi}{\sqrt{a^{2}-b^{2}}}$.

Claude Leibovici · Answer 4 · 2022-08-13T07:07:24.130

Even the antiderivative exists $$I_n=\int \frac{d x}{\left(1+\sin ^{2} (x)\right)^{n}}$$ $$I_n=-\frac{\sqrt{-\sin ^2(x)} \sqrt{\cos ^2(x)} \csc (x) \sec (x) }{2 \sqrt{2} (n-1)\,\left(1+\sin^2(x)\right)^{n-1} }\times $$ $$F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{1}{2} \left(1+\sin ^2(x)\right),1+\sin ^2(x)\right)$$

where appears the Appell hypergeometric function of two variables.

For $x\in\big[ 0,\frac \pi 2\big]$ $$I_n=-i \,\, \frac{F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{1+\sin ^2(x)}{2} ,1+\sin ^2(x)\right)}{2 \sqrt{2} (n-1)\,\left(1+\sin ^2(x)\right)^{n-1} }$$

Using the given bounds, the result write again in terms of the Gaussian hypergeometric function.

For $$J_n=\int_0^{\frac \pi 2} \frac{d x}{\left(1+\sin ^{2} (x)\right)^{n}}=\frac{\pi }{2 \sqrt{2}}\,\,\frac{a_n}{b_n}$$ where tha $a_n$ form the (unknown ?) sequence $$\{1,3,19,63,867,3069,22199,81591,2428451,9119601,68993757,\cdots\}$$ and the $b_n$ correspond to sequence $A123854$ in $OEIS$.

Edit

At $x=3$

$$\frac{\partial^{n}}{\partial x^{n}}\left(\frac{1}{\sqrt{x^{2}-1}}\right)$$ satisfy the recurrence relation $$(n+1)\, u_{n}+3 (2 n+3)\, u_{n+1}+8 (n+2)\,u_{n+2}=0$$ with $$u_0=\frac{1}{2 \sqrt{2}}\qquad \text{and} \qquad u_1=-\frac{3}{16 \sqrt{2}}$$

Interesting integral $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}$

4 Answers4