Help with integral from Apostol Calculus

[emjay66]

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.
<br /> \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)<br />

Homework Equations

N/A

The Attempt at a Solution

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

 

[STEMucator]

Perhaps you could write the integral as:

$\int cos^m(x)sin^{-n}(x) dx$ and apply an existing reduction formula.

 

[lurflurf]

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.