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Happy New Year $2024$ everyone!

This is my first post on this site.

I was attempting this question:

$$ \int_0^{\frac{\pi}{2}} \frac{1}{\cos^6x+\sin^2x} dx$$

I am relatively new to advanced calculus so I could only think of a couple of things to do with this.

First try: substitute $\sin^2x =t$. Didn't quite lead me anywhere, since the integrand now got even weirder.

Second try: Try to substitute $\tan x =t$ and simplify the integrand.

Doing that, if I am not wrong, the expression becomes:

$$\int_0^{\infty} \frac{(t^2+1)^2}{t^2(t^2+1)^2+1}dt = \int_0^{\infty} \frac{t^4+2t^2+1}{t^6+2t^4+t^2+1} dt$$

I was unable to do anything after this. Someone please help me compute this integral. Thanks in advance!

  • try tangent half angle substitution – tryst with freedom Dec 31 '23 at 18:24
  • No, you are not wrong. You do get that integral. – José Carlos Santos Dec 31 '23 at 18:31
  • @JoséCarlosSantos I did get that. But how should I proceed further? –  Dec 31 '23 at 18:34
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    I have no idea. Where did you get this question from? It is not the kind of exercise someone you give to a person who is “relatively new to calculus”. – José Carlos Santos Dec 31 '23 at 18:35
  • @JoséCarlosSantos A friend gave it to me as a challenge when I asked him for a "hard" integral. –  Dec 31 '23 at 18:41
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    I suspect that this integral is beyond “hard”. – José Carlos Santos Dec 31 '23 at 18:42
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    I think that $\int_0^{\infty} \frac{t^4+2t^2+1}{t^6+2t^4+t^2+1} , dt$ can be rewrite the integrand as a partial fraction. First, factor the denominator: $t^6+2t^4+t^2+1 = (t^2+1)^2(t^2+t+1)(t^2-t+1) $. Hence It can be express the integrand as the sum of partial fractions. – Sebastiano Dec 31 '23 at 18:51
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    Are you familiar with contour integration? – KStarGamer Dec 31 '23 at 18:57
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    @Sebastiano Your attempted factorisation of the denominator is of degree $8$. It clearly shouldn't be divisible by $t^2+1$, squared or otherwise. – J.G. Dec 31 '23 at 19:00
  • @J.G. I have delete my hint. – Sebastiano Dec 31 '23 at 19:01
  • @KStarGamer No, I am not. –  Dec 31 '23 at 19:18
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    Seconding contour methods. Your integral is also equivalent to$$\frac\pi2 + \int_{-\infty}^\infty \frac{z \left(z^2+1\right)^2 \left(2z^4-2z^2-1\right)}{\left(z^6+z^4+2z^2+1\right)^2} \arctan z , dz$$($y=\cos x$, int. by parts, $z=\frac y{\sqrt{1-y^2}}$) which can be evaluated in a similar manner as shown here. Granted, this is far more work than necessary, your integral is significantly more residue-friendly. – user170231 Dec 31 '23 at 19:23

3 Answers3

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Rewrite the integral as $$I=\int_0^{\frac{\pi}{2}} \frac{1}{\cos^6x + \sin^2x} dx = \int_0^{\frac{\pi}{2}} \frac{1}{\cos^6x - \cos^2x+1} dx $$ Then, utilize the factorization
$$x^3-x+1 =(x-a)(x-b)(x-b^*) $$ with $b+b^* =-a$, $bb^*=-\frac1a$ $$a = -\sqrt[3]{\frac12-\frac16\sqrt{\frac{23}3}} -\sqrt[3]{\frac12+\frac16\sqrt{\frac{23}3}}\approx -1.3247 $$ to partially fractionalize the integrand \begin{align} I=& \ \frac a{2a-3}\int_0^{\frac{\pi}{2}} \frac{dx}{\cos^2x -a}+ \Re \frac2{(b-b^*)(b-a)}\int_0^{\frac{\pi}{2}} \frac{dx}{\cos^2x -b} \\ =& \ \frac{\pi}{2a(2a-3)}-\Re \frac{\pi}{b^2(b-b^*)(b-a)}\\ =& \ \frac{\pi}{2a(2a-3)}+\frac{\pi(1+a^2)a^3}{2(2a-3)} =\frac\pi2+\frac{\pi}{a(2a-3)} \end{align} where $a^3=a-1$ is used to simplify the close-form in the last step.

Quanto
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    Note that, using Bezout's identity, you can write the answer as $\frac{\pi} {46} (10 a^2 - 8 a + 1)$ if you'd like the final answer to be a polynomial in $a$. – Andrew Dec 31 '23 at 22:40
  • @Andrew Thanks. The polynomial version is nicer. – Quanto Dec 31 '23 at 23:50
  • @Quanto Thanks for such an easy-to-understand answer! But can you please show how did you obtain the value for $a$ if possible? –  Jan 01 '24 at 05:15
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    @Irene The analytic solution $a$ is given by the Cardano's formula of the cubic equation. – Quanto Jan 01 '24 at 12:16
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Maple solves this (apparently by solving a cubic equation with Cardano's formula). The final answer:

(61872461*Pi*(3^(1/2)+22345707/61872461*23^(1/2))*2^(2/3)*(3*3^(1/2)*23^(1/2)+ 25)^(1/3)+15374971*(3^(1/2)+5552787/15374971*23^(1/2))*Pi*2^(1/3)*(3*3^(1/2)*23 ^(1/2)+25)^(2/3)+Pi*(222815375*3^(1/2)+80471457*23^(1/2)))/(482828742*23^(1/2)+ 1336892250*3^(1/2))

$$ {\frac {\big[61872461\,\pi\, \left( \sqrt {3}+{\frac {22345707\,\sqrt {23} }{61872461}} \right) {2}^{2/3}\sqrt [3]{3\,\sqrt {3}\sqrt {23}+25}+ 15374971\, \left( \sqrt {3}+{\frac {5552787\,\sqrt {23}}{15374971}} \right) \pi\,\sqrt [3]{2} \left( 3\,\sqrt {3}\sqrt {23}+25 \right) ^{ 2/3}+\pi\, \left( 222815375\,\sqrt {3}+80471457\,\sqrt {23} \right) \big]}{ 482828742\,\sqrt {23}+1336892250\,\sqrt {3}}} $$

$$ = \frac{3\pi}{2}\left(2^{2/3}\left(173+21\sqrt{69}\right)^{1/3}-10\cdot 2^{1/3}\left(173+21\sqrt{69}\right)^{-1/3}-7\right)^{-1}.$$

GEdgar
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    FWIW, this can be more clearly written as $$\frac{3\pi}{2}\left(2^{2/3}\left(173+21\sqrt{69}\right)^{1/3}-10\cdot 2^{1/3}\left(173+21\sqrt{69}\right)^{-1/3}-7\right)^{-1}.$$ – KStarGamer Dec 31 '23 at 20:00
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Using your approach $$I=\int_0^\infty \frac{\left(t^2+1\right)^2}{t^6+2 t^4+t^2+1}\,dt$$ Write $$t^6+2 t^4+t^2+1=(t^2-a)(t^2-b)(t^2-c)$$ where $a$ is a negative real number and $(b,c)$ complex conjugate.

Using partial fraction decomposition, the integrand write $$\frac{(a+1)^2}{(a-b) (a-c) \left(t^2-a\right)}-\frac{(b+1)^2}{(a-b) (b-c) \left(t^2-b\right)}+\frac{(c+1)^2}{(a-c) (b-c) \left(t^2-c\right)}$$ and we face three simple integrals.

This gives $$I=\frac{\pi (a+1)^2}{2 \sqrt{-a} (a-b) (a-c)}+i \frac{\pi}{2} \left(\frac{(b+1)^2}{\sqrt{b} (b-a) (b-c)}+\frac{(c+1)^2}{\sqrt{c} (c-a) (c-b)}\right)$$

Using the hyperbolic solution for the cubic equation $$a=-\frac{4}{3} \cosh ^2\left(\frac{1}{6} \cosh ^{-1}\left(\frac{25}{2}\right)\right)$$ $$(b,c)=-\frac{1}{2} \left(a+2\pm\sqrt{\frac{a (a+2)^2+4}{a}}\right)$$

$$\frac{\pi (a+1)^2}{2 \sqrt{-a} (a-b) (a-c)}+i \frac{\pi}{2} \left(\frac{(b+1)^2}{\sqrt{b} (b-a) (b-c)}+\frac{(c+1)^2}{\sqrt{c} (c-a) (c-b)}\right)$$

Edit

Much inspired by @Quanto's neat solution, I performed as many simplifications as I could to obtain by the end $$\color{blue}{I=\frac \pi 2 \left(1+\frac{3\, \text{sech}(t)}{4 \cosh (t)+3 \sqrt{3}} \right)}\quad \text{where}\quad \color{blue}{t=\frac{1}{3} \cosh ^{-1}\left(\frac{3 \sqrt{3}}{2}\right)}$$

Claude Leibovici
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  • Thank you for your answer, @ClaudeLeibovici! It is really good to know that the closed form of the integral is so nice! This has been a satisfying experience! –  Jan 01 '24 at 12:37