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!
Best Answer
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.