1. ## Another integration problem

If $\displaystyle \displaystyle I_n=\int\frac{\cos^{2n} x}{\sin x} dx$, write down a similar expression for $\displaystyle I_{n+1}$. Hence, by using the identity $\displaystyle \cos^2 x+\sin^2 x\equiv 1$, prove that $\displaystyle (2n+1)I_{n+1}=(2n+1)I_n+\cos^{2n+1} x$. Hence or otherwise determine $\displaystyle \displaystyle\int\frac{\cos^6 x}{\sin x} dx$

I found $\displaystyle I_{n+1}=\frac{\cos^{2n+2}}{\sin x} dx$
But I don't know how to do the next part.
Thanks!

2. $\displaystyle \displaystyle{I_n = \int{\frac{\cos^{2n}{x}}{\sin{x}}\,dx}}$.

Then $\displaystyle \displaystyle{I_{n+1} = \int{\frac{\cos^{2(n+1)}{x}}{\sin{x}}\,dx}}$

$\displaystyle \displaystyle{= \int{\frac{\cos^{2n+2}{x}}{\sin{x}}\,dx}}$

$\displaystyle \displaystyle{= \int{\cos^2{x}\,\frac{\cos^{2n}{x}}{\sin{x}}\,dx}}$

$\displaystyle \displaystyle{= \int{(1 - \sin^2{x})\,\frac{\cos^{2n}{x}}{\sin{x}}\,dx}}$

$\displaystyle \displaystyle{= \int{\frac{\cos^{2n}{x}}{\sin{x}} - \sin{x}\cos^{2n}{x}\,dx}}$

$\displaystyle \displaystyle{= \int{\frac{\cos^{2n}{x}}{\sin{x}}\,dx} - \int{\sin{x}\cos^{2n}{x}\,dx}}$

$\displaystyle \displaystyle{= \int{\frac{\cos^{2n}{x}}{\sin{x}}\,dx} - \int{u^{2n}\,du}}$ after making the substitution $\displaystyle \displaystyle{u = \cos{x}}$

$\displaystyle \displaystyle{= \int{\frac{\cos^{2n}{x}}{\sin{x}}\,dx} - \frac{u^{2n + 1}}{2n + 1}}$

$\displaystyle \displaystyle{= \int{\frac{\cos^{2n}{x}}{\sin{x}}\,dx} - \frac{\cos^{2n + 1}{x}}{2n+1}}$

$\displaystyle \displaystyle{= I_n - \frac{\cos^{2n + 1}{x}}{2n + 1}}$.

Since we have

$\displaystyle \displaystyle{I_{n + 1} = I_n - \frac{\cos^{2n + 1}{x}}{2n + 1}}$

$\displaystyle \displaystyle{(2n + 1)\,I_{n + 1} = (2n + 1)\left(I_n - \frac{\cos^{2n+1}{x}}{2n + 1}\right)}$

$\displaystyle \displaystyle{(2n + 1)\,I_{n + 1} = (2n + 1)\,I_n - \cos^{2n + 1}{x}}$.

Now since you're trying to find $\displaystyle \displaystyle{\int{\frac{\cos^6{x}}{\sin{x}}\,dx}}$, this is $\displaystyle I_6$.

Then you have

$\displaystyle \displaystyle{I_6 = I_5 - \frac{\cos^{11}{x}}{11}}$.

This means you will need to find $\displaystyle I_5, I_4, I_3, I_2$ and $\displaystyle I_1$.