Integration by parts, Reduction 
I was able to complete part (a) easily by using integration by parts. I ended up getting: 
$$I(n) = -\frac{1}{n} \cos x\cdot \sin^{n-1}x + \frac{n-1}{n}· I(n-2)$$
For question (b), When I integrated $1/\sin^4x$ and subbed in $n = -4$, I get the following equation:
$$\frac{1}{4}·\cos x·\sin^{-5}x + \frac{5}{4} \int sin^{-6}x dx$$
My question is, how do I integrate $\sin^{-6}x$ because it's not the same as integrating $\sin^{6}x$, which will actually get you somewhere. It feels like I'm going in a loop when integrating $\sin^{-6}x$.
I might have went wrong somewhere, help would be very much appreciated :)

Putting $n=-2,$ in  $$I_n=-\frac1n\cos x\sin^{n-1}x+\frac{n-1}nI_{n-2}$$
we get   $$I_{-2}=-\frac1{(-2)}\cos x\sin^{-2-1}x+\frac{(-2-1)}{(-2)}I_{-2-2}$$
$$\implies  \frac32I_{-4}=I_{-2}-\frac{\cos x}{2\sin^3x}$$
Now, $$I_{-2}=\int\sin^{-2}xdx=\int \csc^2xdx=-\cot x+C$$
Can you finish it from here?
