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 :)

Here is a start,
$$ I_n = \int \sin^{n-2}(x) \sin^2(x) dx = \int \sin^{n-2}(x) (1-\cos^2(x)) dx $$
$$ = \int \sin^{n-2}(x)dx - \int \sin^{n-2}(x)\cos(x)\cos(x)dx \dots. $$
Can you finish it? Use integration by parts to evaluate the last integral.
