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?
 A: Make the substitution $\cos(x)=u$ to get:
$$\int\frac{7u^2-1}{u^2(1-u^2)^4}du$$
A: 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
A: 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?
A: 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} $
