# Trying to prove trigonometric integrals on a quarter of circle

1. Mar 22, 2016

### supermiedos

1. The problem statement, all variables and given/known data
I want to prove that:

2. Relevant equations

3. The attempt at a solution
I tried using the trigonometric identity:
sen2x = senx cosx / 2, so, I got:

1/2m∫(sen2x)mdx, x from 0 to pi/2, but now I don't know how to proceed. Can you help me please?

2. Mar 22, 2016

### Staff: Mentor

Is m an integer? Is m a positive integer? These should be stated as part of the problem statement. Assuming that m is a nonnegative integer, I think you need to break up the problem into two cases: 1) m is an even integer; 2) m is an odd integer.

For the first case (m is even), convert all of the sine factors on the left side to cosines, using $sin^2(x) = 1 - cos^2(x)$. At that point you can use your double angle formula.
For the second case (m is odd), convert all but one of the sine factors on the left side to cosines, as above. The left-over sine factor can be used in an ordinary substitution.

Last edited: Mar 22, 2016
3. Mar 22, 2016

### supermiedos

Thank you for your response. As you said, the problem should have stated the "nature" of m. I tried first using your suggestion when m is odd. This is my procedure, but I got stuck:

But I don't know what to do next.

Last edited by a moderator: Mar 22, 2016
4. Mar 22, 2016

### Staff: Mentor

Ah, the perils of giving advice when you haven't worked the problem...

Your original thought might be the way to go, since $\cos^m(x) \sin^m(x) dx = (\cos x \cdot \sin x)^m = \frac 1 {2^m} \sin^m(2x)$. Then, if you can show that the integral of this function equals the integral on the right side of what you're trying to prove, then you're done. At the moment, I don't see how to do this, but I'll give it some thought.

5. Mar 22, 2016

### supermiedos

Hehe, don't worry. I was trying to use geometric arguments (showing that the area below sin^m(2x) equals the area belos cos^m(x), from 0 to pi/2). I can't see the way, also. Thank you for your help

6. Mar 22, 2016

### Staff: Mentor

I think two applications of integration by parts on $\int \sin^m(2x)dx$ might be fruitful. I'm giving this a shot right now

7. Mar 22, 2016

### vela

Staff Emeritus
It seems like you should be able to use some sort of symmetry argument to change the sine into a cosine in your last integral.

Perhaps try something like $\cos x = \sin(x+\frac{\pi}{2})$.

Last edited: Mar 22, 2016
8. Mar 23, 2016

### LCKurtz

@supermiedos: After applying the double angle formula, you are wanting to show$$\int_0^{\frac {\pi}2}\sin^m(2x)~dx=\int_0^{\frac {\pi}2}\cos^m(x)~dx$$
Let $u=2x$ in the left integral, we get$$\int_0^{\frac {\pi}2}\sin^m(2x)~dx=\frac 1 2 \int_0^{\pi}\sin^m(u)~du$$Since the sine function is symmetric about $\frac{\pi} 2$ on the interval $[0,\pi]$, we have$$\int_0^{\pi}\sin^m(u)~du = 2\int_0^{\frac{\pi} 2}\sin^m(u)~du$$Putting this together gives$$\int_0^{\frac {\pi}2}\sin^m(2x)~dx=\int_0^{\frac{\pi} 2}\sin^m(u)~du=\int_0^{\frac{\pi} 2}\sin^m(x)~dx$$
And, finally, by symmetry$$\int_0^{\frac{\pi} 2}\sin^m(x)~dx=\int_0^{\frac {\pi}2}\cos^m(x)~dx$$
That should do it unless I've overlooked something. (I know, I'm taking a chance by assuming this isn't actually a homework problem).

9. Mar 23, 2016

### supermiedos

I tried it, but gotten nowhere

10. Mar 23, 2016

### supermiedos

Thank you, I also tried it, but I got stuck

11. Mar 23, 2016

### supermiedos

Amazing! Thank you! I understood all steps, except one. How can I prove your last statement? Using symetry?

And don't worry, it's not homework. I'm self studying mathematics :)

12. Mar 23, 2016

### vela

Staff Emeritus
You can use the identity to show that
$$\int_0^\pi \sin^m u\,du = \int_{-\pi/2}^{\pi/2} \cos^m u\,du$$ and then use the symmetry of cosine to get the integral you want.

13. Mar 23, 2016

### LCKurtz

Use $\sin x = \cos(\frac {\pi} 2 - x)$ and do the obvious $u$ substitution.

14. Mar 24, 2016

### supermiedos

Thank you, both of you. I got it :)