Indefinite integral of $\cos^2(x)/\sin^6(x)$

Question

I can't solve this integral, can anybody help me ?

$$\int\frac{\cos^2x}{\sin^6x}\mathrm dx$$

I've tried with $\frac{\cos2x+1}{2}=\cos^2x$ transformations but still cannot get the result.

Answer 1

$\displaystyle \dfrac{\cos^2x}{\sin^6 x}=\csc^4 x \cot^2 x= (\cot x+\cot^3 x)^2 \csc^2 x$

Now integrating and letting $\cot x=y$

$I=-\int (y^2+y^6+2y^4)\; dy$

Now you can proceed.

Answer 2

If you absolutely want to avoid trig, here you go
$$\int\frac{\cos^2x}{\sin^6x}dx$$ $$\int\frac{1-\sin^2x}{\sin^6x}dx$$ We now let $u = \sin^2(x) \implies u' = 2\sqrt{u-u^2}$
This shrinks our interval a bit since the square root actually gives the absolute value, so we need really have the above for $[0+\pi n,\pi/2+\pi n]$. $$\frac{1}{2}\int\frac{1-u}{u^6\sqrt{u-u^2}}du$$ You can now multiply out the denominator, make a substitution to move the square root to the top, and simplify. Good luck.

Answer 3

$$ \begin{aligned} \int \frac{\cos ^2 x}{\sin ^6 x} d x & =\int \cot ^2 x \csc ^4 x d x \\ & =-\int \cot ^2 x\left(1+\cot ^2 x\right) d(\cot x) \\ & =-\frac{\cot ^3 x}{3}-\frac{\cot ^5 x}{5}+C \end{aligned} $$