# 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 :)

• Maybe you should integrate $\sin^{-2}x$. May 5, 2013 at 12:44
• Start with $n=-2$ May 5, 2013 at 12:44
• ohh okay, i think i get it now May 5, 2013 at 12:50

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?

• how do i ask a new question? May 6, 2013 at 14:09
• @GeorgeRandall, press "ASK QUESTION" at the top right May 6, 2013 at 15:20

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.

• George "was able to complete part (a) easily " May 5, 2013 at 13:01
• @labbhattacharjee: I just paid attention. We can just leave it for those who may not know how to find part (a). Thanks for the comment. May 5, 2013 at 13:06