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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 :)

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

2 Answers 2

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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?

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

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  • $\begingroup$ George "was able to complete part (a) easily " $\endgroup$ May 5, 2013 at 13:01
  • $\begingroup$ @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. $\endgroup$ May 5, 2013 at 13:06

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