Integrate $ \int \frac{dx}{\cos^3x-\sin^3x}$

Question

So I tried expressing all of it in terms of $\tan\frac x2$, and substitute it, but it became way too long and complicated. I am looking for the approach which would require the least number of steps to get the final answer.

Answer 1

The integral is not simple.

The answer is

$$ \dfrac{2\arctan\left(\left(\sqrt{3}+1\right)\tan\left(\frac{x}{2}\right)+1\right)-2\arctan\left(\left(\sqrt{3}-1\right)\tan\left(\frac{x}{2}\right)-1\right)+\sqrt{2}\left(\ln\left(\left|\tan\left(\frac{x}{2}\right)+\sqrt{2}+1\right|\right)-\ln\left(\left|\tan\left(\frac{x}{2}\right)-\sqrt{2}+1\right|\right)\right)}{3} $$

which is certainly not pretty.

Hence,

1. The solution will be complicated no matter what.
2. Online integral solvers may help you with steps.
3. Interesting insights may be provided here in the answers, but you will surely not get a short solution.

Answer 2

Hint:

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Multiply numerator & denominator by $\cos x-\sin x$

As $\int(\cos x-\sin x)dx=\sin x+\cos x,$

Choose $\sin x+\cos x=u\implies2\sin x\cos x=u^2-1$

and $(\cos x-\sin x)^2=2-u^2$