Evaluating this integral : $ \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx $ The question :
$$ \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx $$
I tried dividing by $\cos^2 x$ and splitting the fraction.
That turned out to be complicated(Atleast for me!)
How do I proceed now?

The integration is $$\int \frac{dx}{\sin^7x\cos^2x}-\int\csc^7xdx$$
Using this repeatedly,  $$\frac{m-1}{n+1}\int\sin^{m-2}\cos^n dx=\frac{\sin^{m-1}x\cos^{n+1}x}{m+n}+\int \sin^mx\cos^n dx,$$
$$\text{we can reach from }\int \frac{dx}{\sin^7x\cos^2x}dx\text{ to } \int \frac{\sin xdx}{\cos^2x}dx$$ 
Now use the Reduction Formula of $\int\csc^nxdx$ for the second/last integral
