prove that:
$$\int {\cos^{m}x\over \sin^{n}x}dx = {\cos^{m-1}x\over (m-n)\sin^{n-1}x} + {m-1\over m-n}\int {\cos^{m-2}x\over \sin^{n}x}dx +C$$
What I did:
$$\int \cos^{m}x\sin^{-n}xdx= \int \cos x\cos^{m-1}x\sin^{-n}xdx$$
then I used integration by parts:
$u=\cos^{m-1}x$ $$du=-(m-1)\cos^{m-2}x\sin xdx$$ $$dv=\cos x\sin^{-n}xdx$$ and $$v={\sin^{-n+1}x\over-n+1}$$
hence:
$$\int {\cos^{m}x\over \sin^{n}x}dx= -{\cos^{m-1}x\over (n-1)\sin^{n-1}x} – {m-1\over n-1}\int {\cos^{m-2}x\over \sin^{n-2}x}dx$$
So I would like you to tell me what can I do to solve this and where is my mistake
Best Answer
So far so good. When solving a problem like this, you have to look at what you should prove. Now you have got to: $\displaystyle\int\dfrac{\cos^m x}{\sin^n x} dx = -\dfrac{\cos^{m-1}x}{(n-1)\sin^{n-1}x} - \dfrac{m-1}{n-1}\displaystyle\int \dfrac{\cos^{m-2} x}{\sin^{n-2} x} dx$.
Now basically, you have to manipulate $\dfrac{\cos^{m-2} x}{\sin^{n-2} x}$, to get $\dfrac{\cos^{m-2} x}{\sin^{n} x}$ as required by the problem. You can start by multiplying both numerator, and denominator by $\sin^2 x$, like this:
$$\begin{align}\displaystyle\int \dfrac{\cos^{m-2} x}{\sin^{n-2} x} dx &= \displaystyle\int \dfrac{\cos^{m-2} x\sin^2x}{\sin^{n} x} dx \\ &= \displaystyle\int \dfrac{\cos^{m-2} x(1-\cos^2x)}{\sin^{n} x}dx \\ &= \displaystyle\int \dfrac{\cos^{m-2} x-\cos^mx}{\sin^{n} x}dx \end{align}$$
I think you can go from here, right? :)
This reminds me of a problem in a grade 12 math book in my country.