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?

Integrating by parts,
$$\int \frac{dx}{\sin^7x\cos^2x}=\int \csc^7x\sec^2xdx=\csc^7x\int \sec^2xdx-\int\left( \frac{d\csc^7x}{dx}\cdot \sec^2xdx \right)dx$$
$$=\csc^7x\cdot\tan x-\int\left(7\csc^6x(-\csc x\cot x)\tan x\right)dx$$
$$=\csc^7x\cdot\frac{\sin x}{\cos x}+7\int \csc^7xdx$$
$$=\csc^6x\cdot \sec x+7\int\frac{dx}{\sin^7xdx}$$
Can you take it from here?
