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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?

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

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What will be the next step? This doesn't seem trivial to me. – Inceptio May 8 '13 at 15:54
@Inceptio Partial fractions ${}{}{}{}$ – Amr May 8 '13 at 15:55
Does it yield the answer? There might be a easier way, right? – Inceptio May 8 '13 at 15:56
@Inceptio The other option is to use reduction formulae to find the last integral – Amr May 8 '13 at 15:59
@VijayRaghavan I think you forgot to add the link – Amr May 8 '13 at 16:02

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

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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?

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@Inceptio, how about this one? – lab bhattacharjee May 8 '13 at 16:20
The comment really didn't notify me. This is more direct. (+1) – Inceptio May 11 '13 at 8:33

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