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$$\int \sin ^m\left(x\right)\cos ^n\left(x\right)$$ m,n are natural numbers and I'm asked to find the recurrent formula for this integral.

Now I know there are three cases to look at namely: $$ m>n$$ $$ m<n$$ $$ m=n$$

The last case I'm able to solve, but I'm not sure about the first two. I'm using integration by parts so for example if we take the first case: $$u=\sin ^m\left(x\right)\cos ^{n-1}\left(x\right)$$ $$dv\:=\:\int \cos \left(x\right)$$

Did I choose the wrong u and dv, because I am not able to solve it. At one point I get something like this: $\int \sin ^m\left(x\right)\cos ^{n-2}\left(x\right)$ in the expression and that's quite problematic.

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    $\begingroup$ This is how this integral proceeds though, that integral you obtain is simply $I(m, n-2)$, so now you have a recurrence relation between $I(m,n)$ and the next integral down $\endgroup$ – Triatticus Feb 29 '16 at 20:44
  • $\begingroup$ So I was correct and it's alright to write it as I(m,n-2)? $\endgroup$ – gvidoje Feb 29 '16 at 21:28
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Integration by parts yields:

$$\int\sin^m(x)\cos^n(x)dx=-\frac1{m+1}\sin^{m+1}(x)\cos^{n+1}(x)-\frac{n+1}{m+1}\int\sin^{m+2}(x)\cos^{n}(x)dx$$

Turning this around, you get:

$$\frac{n+1}{m+1}\int\sin^{m+2}(x)\cos^{n}(x)dx=-\left(\frac1{m+1}\sin^{m+1}(x)\cos^{n+1}(x)+\int\sin^m(x)\cos^n(x)dx\right)$$

$$\int\sin^{m+2}(x)\cos^{n}(x)dx=-\left(\frac1{n+1}\sin^{m+1}(x)\cos^{n+1}(x)+\frac{m+1}{n+1}\int\sin^m(x)\cos^n(x)dx\right)$$

And so, if $m$ is even, the solution will eventually be found. If $m$ is odd, you could use $\int\sin(x)\cos^n(x)dx=-\cos^{n+1}(x)$

I shall attempt at the general solution, but it will be messy.

$$\int\sin^{m+2}(x)\cos^{n}(x)dx=-\frac1{n+1}\sin^{m+1}(x)\cos^{n+1}(x)+\frac{m+1}{(n+1)^2}\sin^{m-1}(x)\cos^{n+1}(x)-\frac{(m+1)(m-1)}{(n+1)^3}\sin^{m-3}(x)\cos^{m+1}(x)\dots$$

$$\int\sin^{m+2}(x)\cos^{n}(x)dx=\cos^{m+1}(x)\sum_{i=0}^{\infty}\frac{(-1)^{i+1}(m+i)!}{m!(n+1)^{i+1}}\sin^{m-2i+1}(x)$$

Only use the summation until $m-2i+1=0,1$, then evaluate the last remaining integral. I don't believe the summation converges.

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I may have a different approach, provided that you know some cute things like Binomial Expansion and fractional summations.

Let's start with this (this will work for every $n$ and $m$):

$$z = \sin(x) ~~~~~ \text{d}z = \cos(x) = \sqrt{1 - \sin^2(x)} = \sqrt{1 - z^2}\ \text{d}x$$

Thus

$$\text{d}x = \frac{\text{d}z}{\sqrt{1 - z^2}} ~~~~~~~ \cos^n(x) = [\cos(x)]^n = \left(\sqrt{1 - \sin^2(x)}\right)^n = (1 - z^2)^{n/2}$$

Substituting into the integral and we get

$$\int z^m(1 - z)^{n/2}\frac{\text{d}z}{\sqrt{1 - z^2}} = \int z^m(1 - z^2)^{n/2 - 1/2}\ \text{d}z$$

Let's not call for simplicity $n/2 - 1/2 = \alpha$. Now we use the Binomial Expansion:

$$(1 - z^2)^{\alpha} = \sum_{k = 0}^{\alpha} \binom \alpha k (-z^2)^k\ \text{d}z$$

Thus substituting we get:

$$\sum_{k = 0}^{\alpha} \binom \alpha k (-1)^k\int z^{m+2k}\ \text{d}z$$

A trivial integration will lead us to

$$\sum_{k = 0}^{\alpha} \binom \alpha k (-1)^k \frac{z^{m+2k+1}}{m+2k+1}$$

Now substituting back for $z = \sin(x)$, and $\alpha = \frac{n-1}{2}$ we have

$$\boxed{\sum_{k = 0}^{\frac{n-1}{2}} \binom{\frac{n-1}{2}}{ k} (-1)^k \frac{\sin^{m+2k+1}(x)}{m+2k+1}}$$

Simple proof

Supposing we have $n = m = 1$, namely we are integrating $\int\sin(x)\cos(x)\ \text{d}x$ then we have

$$\frac{n-1}{2} = 0 ~~~~~~~ m + 2k + 1 = 1 + 0 + 1 = 2$$

thence

$$\int\sin(x)\cos(x)\ \text{d}x = \sum_{k = 0}^{0} \binom{0}{0} (-1)^0 \frac{\sin^{2}(x)}{2} = \frac{\sin^2(x)}{2}$$

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    $\begingroup$ Darn it, I was going to post this answer in my answer today! Nice one. $\endgroup$ – Simply Beautiful Art Mar 1 '16 at 12:28
  • $\begingroup$ @SimpleArt lol! Hey you did a great work too!! :D $\endgroup$ – Turing Mar 1 '16 at 13:17
  • $\begingroup$ ;D Thanks. ${}$ $\endgroup$ – Simply Beautiful Art Mar 1 '16 at 13:18

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