Find this integral $I=\int\frac{1}{\sin^5{x}+\cos^6{x}}dx$

Question

How find this $$I=\int\dfrac{1}{\sin^5{x}+\cos^6{x}}dx$$

this problem is from china exam.so I think have closed form

I find this [wolf]

[wolf]:http://www.wolframalpha.com/input/?i=1%2F%28%28sinx%29%5E5%2B%28cosx%29%5E6%29dx enter image description here

such this example: use wolf give us too hard form,But I can use hand to solve it:

How find this integral $I=\int\frac{1}{\sin^5{x}+\cos^5{x}}dx$

see this wolf

enter image description here

Wolfram's solution is telling you that the solution is in terms of a root of a polynomial which is not solvable by radicals. — Ian, Dec 08 '14 at 16:14
@Dr.SonnhardGraubner,sorry,I can't find link,because china most exam problem don't like post intenet — math110, Dec 08 '14 at 16:21
I'd start by changing $\cos^6x=(1-\sin^2x)^3$ and then do the change of variable $u=\sin x$. I'm not sure this works, but it's a start. — hjhjhj57, Dec 08 '14 at 16:22
@hjhjhj57 that doesn't work because there is no $\cos x$ in the numerator — Dylan, Dec 15 '14 at 01:04

score 1 · Answer 1 · answered Dec 08 '14 at 21:05

The best i can do:

By substituting \begin{align*} t&=\tan\frac x2\\ \sin x&=\frac{2t}{1+t^2}\\ \cos x&=\frac{1-t^2}{1+t^2} \end{align*} the integral can be written as $$2\int\frac{\left(1+t^2\right)^5}{t^{12}-6t^{10}+32t^7-20t^6+32t^5+15t^4-6t^2+1}\,dt$$ This integral is solvable by partial fractions if we know the roots of the denominator, but the roots can probably not be expressed in any nice way. Wolfram Alpha is just calling the roots $\omega$ I believe.

Find this integral $I=\int\frac{1}{\sin^5{x}+\cos^6{x}}dx$

1 Answers1