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I seem to be stuck trying to prove the following integral $$ \int\frac{\cos^mx}{\sin^nx}dx = -\frac{\cos^{m+1}x}{(n-1)\sin^{n-1}x}-\frac{m-n+2}{n-1}\int\frac{\cos^mx}{\sin^{n-2}x} dx + C\,\,(n \neq 1) $$ My thinking so far has been that if I take $$ I = \int\frac{\cos^mx}{\sin^nx}dx $$ I have been able to prove that $$ I = -\frac{\cos^{m-1}x}{(n-1)\sin^{n-1}x} - \frac{m-1}{n-1}\int\frac{\cos^{m-2}x}{\sin^{n-2}x}\,dx+C\,\,\,\,\,(1) $$ and $$ I = \frac{\cos^{m-1}x}{(m-n)\sin^{n-1}x} + \frac{m-1}{m-n}\int\frac{\cos^{m-2}x}{\sin^nx}\,dx+C\,\,\,\,\,(2) $$ but showing that $$ I = -\frac{\cos^{m+1}x}{(n-1)\sin^{n-1}x}-\frac{m-n+2}{n-1}\int\frac{\cos^mx}{\sin^{n-2}x} dx + C $$ 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 $$ I = -\frac{\cos^{m+1}x}{(m+1)\sin^{n+1}x} - \frac{n+1}{m+1}\int\frac{\cos^{m+2}x}{\sin^{n+2}x}\, dx + C\,\,\,\,\,(3) $$ but there's obviously more to it.

Any broad hints would be more than welcome.

emjay
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2 Answers2

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Here is a much simpler answer, based on the work you have done already. Denote your integral by $I_{m,n}$. Then your first identity is $$I_{m,n}=-\frac{\cos^{m-1}x}{(n-1)\sin^{n-1}x} - \frac{m-1}{n-1}I_{m-2,n-2}\ .$$ This implies $$I_{m+2,n}=-\frac{\cos^{m+1}x}{(n-1)\sin^{n-1}x} - \frac{m+1}{n-1}I_{m,n-2}\ .$$ Now we have $$I_{m,n}=\int\frac{\cos^mx}{\sin^nx}(\cos^2x+\sin^2x)\,dx=I_{m+2,n}+I_{m,n-2} \ ,$$ and substituting the previous result into this gives what you want directly.

David
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  • Wow, that's a REALLY elegant answer!!!! Much more subtle than my sledgehammer approach. Much appreciated for that. – emjay Apr 30 '14 at 06:33
  • Thanks! . . . but give yourself credit for obtaining the first identity! – David Apr 30 '14 at 06:49
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After a little more thought (and a lot more sleep) I finally figured out the answer:
Let $$ I_1 = \int\frac{\cos^mx}{\sin^{n-2}x}\, dx = \int\frac{\cos^mx}{\sin^{n-1}x}. \sin x\, dx $$ Integrate $I_1$ by parts, choose $u=\frac{\cos^mx}{\sin^{n-1}x}$, $dv = \sin x\, dx$ so $v = -\cos x$ $$ du = \frac{\sin^{n-1}x(-m\cos^{m-1}x\sin x)-\cos^mx((n-1)\sin^{n-2}x\cos x)}{(\sin^{n-1}x)^2} $$ $$ =\frac{sin^{n-1}x\cos^{m-1}x(-m\sin x)-\sin^{n-1}x\cos^{m-1}x((n-1)\sin^{-1}x\cos^2x)}{(\sin^{n-1}x)^2} $$ $$ = \frac{\cos^{m-1}x}{\sin^{n-1}x}(-m\sin x - (n-1)\frac{\cos^2 x}{\sin x}) $$ So $$ I_1 = -\frac{\cos^{m+1}x}{\sin^{n-1}x} - \int\frac{\cos^{m-1}x}{\sin^{n-1}x}(-m\sin x - (n-1)\frac{\cos^2 x}{\sin x})(-\cos x)\, dx $$ $$ = -\frac{\cos^{m+1}x}{\sin^{n-1}x} - \int\frac{\cos^{m-1}x}{\sin^{n-1}x}(m\sin x\cos x + (n-1)\frac{\cos^3x}{\sin x})\, dx $$ $$ = -\frac{\cos^{m+1}x}{\sin^{n-1}x} - m\int\frac{\cos^mx}{\sin^{n-2}x}\,dx - (n-1)\int\frac{\cos^{m+2}x}{\sin^nx}\, dx $$ which implies $$ (1+m)I_1 = -\frac{\cos^{m+1}x}{\sin^{n-1}x} - (n-1)\int\frac{\cos^{m+2}x}{\sin^nx}\, dx $$ $$ = -\frac{\cos^{m+1}x}{\sin^{n-1}x} - (n-1)\int\frac{\cos^mx. \cos^2 x}{\sin^nx}\, dx $$ $$ = -\frac{\cos^{m+1}x}{\sin^{n-1}x} - (n-1)\int\frac{\cos^mx.(1-\sin^2 x)}{\sin^nx}\, dx $$ $$ = -\frac{\cos^{m+1}x}{\sin^{n-1}x} - (n-1)(\int\frac{\cos^mx}{\sin^nx}\,dx - \int\frac{\cos^mx}{\sin^{n-2}x}\,dx ) $$ Let $$ I = \int\frac{\cos^mx}{\sin^nx}\,dx $$ so we now have $$ I_1 = \frac{-\cos^{m+1}x}{(1+m)\sin^{n-1}x} - \frac{n-1}{m+1}I + \frac{n-1}{m+1}\int\frac{\cos^mx}{\sin^{n-2}x}\,dx $$ $\implies$ $$ (1-\frac{n-1}{m+1})I_1 = \frac{-\cos^{m+1}x}{(1+m)\sin^{n-1}x} - \frac{n-1}{m+1}I $$ $\implies$ $$ \frac{m-n+2}{m+1}I_1 = \frac{-\cos^{m+1}x}{(1+m)\sin^{n-1}x} - \frac{n-1}{m+1}I $$ $\implies$ $$ \frac{n-1}{m+1}I = \frac{-\cos^{m+1}x}{(1+m)\sin^{n-1}x} - \frac{m-n+2}{m+1}I_1 $$ $\implies$ $$ I = \frac{m+1}{n-1}.\frac{-\cos^{m+1}x}{(1+m)\sin^{n-1}x} - \frac{m+1}{n-1}.\frac{m-n+2}{m+1}I_1 $$ $$ = \frac{-\cos^{m+1}x}{(n-1)\sin^{n-1}x} - \frac{m-n+2}{n-1}I_1 $$ All comments greatly appreciated!

emjay
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