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?

Let $\displaystyle I = \int\frac{1-7\cos^2 x}{\sin^7 x\cos^2 x} = \int\frac{\sin^2 x-6\cos^2 x}{\sin^7 x\cos^2 x}dx$
$\displaystyle I = \int\frac{\sin^7 x-6\sin^5 x\cos^2 x}{\sin^{12}x\cos^2 x}dx =  -\int \bigg[\frac{1}{(\sin^ 6 x\cos x)^2}\bigg]'dx = -\frac{1}{\sin^6 x\cos x}+\mathcal{C} $
