Proving Equality of Definite Integrals with Trigonometric Functions

[AndersCarlos]

Homework Statement

I've been solving a problem, the solution is complete, however, I must prove that the following relation is true:

\int_{0}^{\frac{\pi}{2}} sin^m 2x dx = \int_{0}^{\frac{\pi}{2}} cos^m x dx
for any m.

Homework Equations

-

The Attempt at a Solution

Well, I've trying to find some kind of solution by using substitution, however, when I computed both integrals in indefinite form on WolframAlpha, to see if I was following the right path, it showed me an answer that contained a "hypergeometric function", which I haven't learned yet.

 

[micromass]

Try a suitable subsitution. How can you make a sine into a cosine??

 

[AndersCarlos]

micromass:

Using the Pythagorean trigonometric identity.
Well, this would become (if I take the positive root): sin^m 2x = (1-cos^2 2x)^{\frac{m}{2}}

I'm trying 'u' = cos x this time.

 

[micromass]

What is

\sin(\frac{\pi}{2}-x)

??

 

[AndersCarlos]

micromass:

cos x

 

[Curious3141]

AndersCarlos, this is more of the same based on what we discussed yesterday.

Micromass has given you a very big hint. Try to convert \sin 2x into \cos u. What substitution would do that? (Note that your sub must also convert that double angle into a single angle). Hint: There's a \frac{\pi}{4} somewhere in there.

After that, there's more of that "even function" manipulation we were talking about yesterday.

 

[AndersCarlos]

micromass and Curious3141:

Well, I chose that: 2x = π/2 - u
then, dx = -du/2
\int_{0}^{\frac{\pi}{2}} sin^m (2x)dx = - \int_{\frac{\pi}{2}}^{\frac{-\pi}{2}} \frac{sin^m (\frac{\pi}{2} - u)}{2}du = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{cos^m (u)}{2} du = \int_{0}^{\frac{\pi}{2}} cos^m (u) du = \int_{0}^{\frac{\pi}{2}} cos^m (x) dx

Well, I didn't see any π/4 during the process, but if there is anything wrong with this proof, sorry because I wrote it quite fast. Thank you both for your help.

Edit: I have forgotten to put the 'm' exponent through the process, just fixed it.

 

[SammyS]

Looks good !

 

[Curious3141]

micromass and Curious3141:

Well, I chose that: 2x = π/2 - u
then, dx = -du/2
\int_{0}^{\frac{\pi}{2}} sin^m (2x)dx = - \int_{\frac{\pi}{2}}^{\frac{-\pi}{2}} \frac{sin^m (\frac{\pi}{2} - u)}{2}du = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{cos^m (u)}{2} du = \int_{0}^{\frac{\pi}{2}} cos^m (u) du = \int_{0}^{\frac{\pi}{2}} cos^m (x) dx

Well, I didn't see any π/4 during the process, but if there is anything wrong with this proof, sorry because I wrote it quite fast. Thank you both for your help.

Edit: I have forgotten to put the 'm' exponent through the process, just fixed it.

Well, the pi/4 is implicit in your proof. What's x in terms of u?

Anyway, good job.