Question

Integrate:

Answer 1

\[\int\limits dx/(\sin^6x + \cos^6x)\]

Answer 2

I got till
\[\int\limits dx/(1-3\sin^2xcos^2x)\]

Answer 3

What do I do now?

Answer 4

@mukushla

Answer 5

are u sure? im getting\[\sin^6 x+\cos^6 x=1+3 \ \sin^2x(\sin^2x-1)\]

Answer 6

oh sorry...its same

Answer 7

Okay.
I got it.
Hint : take tanx - cotx as t.

Answer 8

I think that would work.
\[I =\frac{ \frac{ dx }{ \sin^2xcos^2x } }{ 1/\sin^2xcos^2x - 1 }\]

\[tanx - cotx = t => \sec^2x + cosec^2x=dt=1/\sin^2xcos^2x=Numerator.\]

Answer 9

or one can write\[\frac{1}{1-3 \ \sin^2 x \ \cos^2 x}=\frac{1}{1-\frac{3}{4} \ \sin^2 (2x)}=\frac{1}{1-\frac{3}{4} \ \frac{1}{1+\cot^2 (2x)}} \]and letting\[u=\cot (2x)\]

Answer 10

Okay.
Thats pretty awesome too.

Answer 11

\[\frac{1}{\sin ^6(x)+\cos ^6(x)}=\frac{8}{3 \cos (4 x)+5} \]
WolframAlpha shows the integration steps for the RHS but not the LHS.