Help with integral from Apostol Calculus

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Help with integral from Apostol "Calculus"

Homework Statement


I seem to be stuck trying to prove the following integral from Apostol "Calculus" Volume 1 Section 5.11, Question 33.
[itex] \int\frac{\cos^mx}{\sin^nx}dx = -\frac{\cos^{m+1}x}{(n-1)\sin^{n-1}x}-\frac{m-n+2}{n-1}\int\frac{\cos^mx}{\sin^{n-2}x} dx + C\,\,(n \neq 1)[/itex]

Homework Equations


N/A

The Attempt at a Solution


My thinking so far has been that if I take
[itex] I = \int\frac{\cos^mx}{\sin^nx}dx[/itex]
I have been able to prove that
[itex] I = -\frac{\cos^{m-1}x}{(n-1)\sin^{n-1}x} - \frac{m-1}{n-1}\int\frac{\cos^{m-2}x}{\sin^{n-2}x}\,dx+C\,\,\,\,\,(1)[/itex]
and
[itex] I = \frac{\cos^{m-1}x}{(m-n)\sin^{n-1}x} + \frac{m-1}{m-n}\int\frac{\cos^{m-2}x}{\sin^nx}\,dx+C\,\,\,\,\,(2)[/itex]
but showing that
[itex] I = -\frac{\cos^{m+1}x}{(n-1)\sin^{n-1}x}-\frac{m-n+2}{n-1}\int\frac{\cos^mx}{\sin^{n-2}x} dx + C[/itex]
seems to be eluding me. I attempted to apply a similar technique what I used on [itex](1)[/itex] to get [itex](2)[/itex] to try to obtain this integral, but it didn't seem to work.
I can also show that
[itex] I = -\frac{\cos^{m+1}x}{(m+1)\sin^{n+1}x} - \frac{n+1}{m+1}\int\frac{\cos^{m+2}x}{\sin^{n+2}x}\, dx + C[/itex]
but there's obviously more to it from this perspective.
Any help would be greatly appreciated.
 
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You need to use $1=\sin^2(x)+\cos^2(x)$

$$\int \! \frac{\cos^{m}(x)}{\sin^{n}(x)} \, \mathrm{d}x=
\int \! \frac{\cos^{m}(x)}{\sin^{n}(x)}(\sin^2(x)+\cos^2(x)) \, \mathrm{d}x=
\\
\int \! \frac{\cos^{m}(x)}{\sin^{n}(x)}\sin^2(x) \, \mathrm{d}x+
\int \! \frac{\cos^{m}(x)}{\sin^{n}(x)}\cos^2(x) \, \mathrm{d}x=
\\
\int \! \frac{\cos^{m}(x)}{\sin^{n-2}(x)} \, \mathrm{d}x+
\int \! \frac{\cos^{m+2}(x)}{\sin^n(x)} \, \mathrm{d}x$$

Integrate the bit with m+2 by parts to reach the desired form.
 
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