1. ## integration question

Had a quiz today, one of the questions was

$\displaystyle \int sin^n(x) ~dx$

:/ I could do it if it were a number for n. I could probably write an algorithm for it if I knew whether it were even or odd, but I don't know how I'm supposed to approach it.

2. ## I would go about saying that

If it is even you rewrite it as $\displaystyle \int[\frac{1-cos(x)}{2}]^{n-2}$ and keep decomposing it using trig identities...this is one where you cant get a definite answer youll get part of an answer followed by another indefinite integral...if I am not mistaken

3. Originally Posted by Mathstud28
If it is even you rewrite it as $\displaystyle \int[\frac{1-cos(x)}{2}]^{n-2}$ and keep decomposing it using trig identities...this is one where you cant get a definite answer youll get part of an answer followed by another indefinite integral...if I am not mistaken
But I don't know the value of n, I don't know if it is even or odd.

4. ## Well I

dont know what to tell you..there is ways of doing it but it involves hypergeometric functions...

5. Originally Posted by angel.white
Had a quiz today, one of the questions was

$\displaystyle \int sin^n(x) ~dx$

:/ I could do it if it were a number for n. I could probably write an algorithm for it if I knew whether it were even or odd, but I don't know how I'm supposed to approach it.
try by parts. you should be able to discern a pattern after doing it two or three times

Let $\displaystyle u = \sin^n x$ and $\displaystyle dv = dx$

6. ## Man

I was just about to say what Jhevon said lol$\displaystyle \int sin(x)^{n},dx$=$\displaystyle xsin(x)^{n}-\int \frac{nxsin(x)^{n}}{tan(x)},dx\$ ad infinitum

7. Originally Posted by Jhevon
try by parts. you should be able to discern a pattern after doing it two or three times

Let $\displaystyle u = \sin^n x$ and $\displaystyle dv = dx$
$\displaystyle x~\sin^nx -\frac n2 x^2~\cos^{n-1}x +$ $\displaystyle \frac {n(n-1)}{1*2*3}x^3\sin^{n-2}x - ... \pm \frac{n(n-1)(n-2)*...*(1)}{1*2*3*...*n}(\sin~ or~ \cos)~x + \frac 1n x^{n-1}+ C$

It feels like there are too many unknowns to say for sure.

8. Originally Posted by angel.white
$\displaystyle x~\sin^nx -\frac n2 x^2~\cos^{n-1}x +$ $\displaystyle \frac {n(n-1)}{1*2*3}x^3\sin^{n-2}x - ... \pm \frac{n(n-1)(n-2)*...*(1)}{1*2*3*...*n}(\sin~ or~ \cos)~x + \frac 1n x^{n-1}+ C$

It feels like there are too many unknowns to say for sure.
indeed. perhaps using the reduction formula would be more fruitful. (it is derived the same way, by the way, using $\displaystyle u = \sin^{n - 1} x$ and $\displaystyle dv = \sin x$)

you could also do two cases, n is odd, or n is even, then you'd know if the last function should be cosine or sine

9. Originally Posted by Jhevon
you could also do two cases, n is odd, or n is even, then you'd know if the last function should be cosine or sine
Yeah, thats what I started to do, but I just got a little overwhelmed (there was huge time crunch on this test too)

10. ## I dont know if this is right but..

since sin(x)≡$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$ woulding $\displaystyle sin(x)^{m}$ or $\displaystyle [sin(x)]^{m}$
be the same as $\displaystyle \bigg[\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\bigg]^{m}$ which is equal to $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m}}{[(2n+1)!]^{m}}$ so $\displaystyle \int\sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m}}{[(2n+1)!]^{m}},dx$=$\displaystyle \sum_{n=0}^{\infty}\int\frac{(-1)^{mn}x^{2mn+mm}}{[(2n+1)!]^{m}},dx$ which is equal to $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{mn}}{[(2n+1)!]^{m}}\int[x^{2nm+m}],dx$=$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m+1}}{[(2n+1)!]^{m}(2mn+m+1)}$

11. Originally Posted by Mathstud28
since sin(x)≡$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$ woulding $\displaystyle sin(x)^{m}$ or $\displaystyle [sin(x)]^{m}$
be the same as $\displaystyle \bigg[\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\bigg]^{m}$ which is equal to $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m}}{[(2n+1)!]^{m}}$ so $\displaystyle \int\sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m}}{[(2n+1)!]^{m}},dx$=$\displaystyle \sum_{n=0}^{\infty}\int\frac{(-1)^{mn}x^{2mn+mm}}{[(2n+1)!]^{m}},dx$ which is equal to $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{mn}}{[(2n+1)!]^{m}}\int[x^{2nm+m}],dx$=$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{mn}x^{2mn+m+1}}{[(2n+1)!]^{m}(2mn+m+1)}$
Wow, thank you for putting all that time and effort in. Unfortunately that is a bit beyond what I have learned, I did not know sin was equal to that sum.

Wouldn't surprise me if we learn that soon, though, we are into sequences and series' now.

12. Originally Posted by angel.white
Yeah, thats what I started to do, but I just got a little overwhelmed (there was huge time crunch on this test too)
Read this. The sine case would be done similarly.

no problem