Integration by Parts - Proving Equations for m and n > 1

[Bazman]

Hi,

[tex] J(m,n) = \int_0^{\frac{\pi}{2}} \cos^m \theta \sin^n \theta d\theta [/tex]


First of all I had to evaluate the following ( I don't know what the correct answers are but here are my calculations:

[tex] J(0,0) = [\theta]_0^{\frac{{\pi}{2}}}=\frac{\pi}{2} [/tex]

[tex] J(0,1) = [-\cos \theta]_0^{\frac{{\pi}{2}}}= 1 [/tex]

[tex] J(1,0) = [ \sin\theta]_0^{\frac{{\pi}{2}}}= 1 [/tex]

[tex] J(1,1) = [\frac{-\cos 2\theta}{4}]_0^{\frac{{\pi}{2}}}= \frac{1}{2} [/tex]

[tex] J(m,1) = [-\frac{\cos^{m+1} \theta}{m+1}]_0^{\frac{{\pi}{2}}}= \frac{1}{m+1} [/tex]

[tex] J(1,n) = [\frac{\sin^{n+1} \theta}{n+1}]_0^{\frac{{\pi}{2}}}= \frac {1}{n+1} [/tex]

Next I am supposed to use integration by parts to prove that for m and n > 1

[tex] J(m,n) = \frac{m-1}{m+n} J(m-2,n) [/tex]

and

[tex] J(m,n) = \frac{n-1}{m+n} J(m,n-2) [/tex]

When I carried out integration by parts I got the following:

taking

[tex] u = \sin^{n-1} \theta [/tex]
[tex] u' = (n-1) \sin^{n-2}\theta \cos \theta [/tex]
[tex] v' = \cos^m \theta \sin \theta [/tex]
[tex] v = -\frac{cos^{m+1}\theta}{m+1} [/tex]

[tex] \frac{-\sin^{n-1} \theta \cos^{m+1}\theta}{m+1} + \int_0^{\frac{\pi}{2}} \frac{n-1}{m+1} \sin^{n-2} \theta \cos^{m+2} \theta d\theta [/tex]

the first term on the RHS equates to zero but the 2nd term is not correct: the denominator and the power to which cos is raised is wrong but I'm not sure how to fix it

 

[VietDao29]

I don't know why, but your LaTeX part is a bit messy for me.
You are correct to the last part. So, you have shown that:
[tex]\int_0 ^ \frac{\pi}{2} \cos ^ m \theta \sin ^ n \theta d( \theta ) = \frac{n - 1}{m + 1} \int_0 ^ \frac{\pi}{2} \cos ^ {\fbox{m + 2}} \theta \sin ^ {\fbox{n - 2}} \theta d( \theta )[/tex]

Note that cos function on the RHS is to the power of m + 2, and you want to prove that:

[tex]\int_0 ^ \frac{\pi}{2} \cos ^ m \theta \sin ^ n \theta d( \theta ) = \frac{n - 1}{m + n} \int_0 ^ \frac{\pi}{2} \cos ^ {\fbox{m}} \theta \sin ^ {\fbox{n - 2}} \theta d( \theta )[/tex].

So what you should do is to split cos^{m + 2}(x) to cos^m(x)cos²(x) = cos^m(1 - sin²(x)), like this:
[tex]RHS = \frac{n - 1}{m + 1} \int_0 ^ \frac{\pi}{2} \cos ^ {m} \theta (1 - \sin ^ 2 \theta) \sin ^ {n - 2} \theta d( \theta ) = \frac{n - 1}{m + 1} \left( \int_0 ^ \frac{\pi}{2} \cos ^ {m} \theta \sin ^ {n - 2} \theta d( \theta ) - \int_0 ^ \frac{\pi}{2} \cos ^ {m} \theta \sin ^ {n} \theta d( \theta ) \right)[/tex].

So we have:

[tex]J(m, n) = \frac{n - 1}{m + 1} \left( J(m, n - 2) - J(m, n) \right)[/tex]
Can you go from here? :)

--------

The other can be done almost the same, except for the first step:
Instead of choosing
u = sin^{n - 1}(x)
and dv = cos^m(x) sin(x) dx, we choose:
u = cos^{m - 1}(x)
and dv = sinⁿ(x) cos(x) dx.

Can you complete the two problems?

 

[Bazman]

Hey VietDao29!

Thanks for your help on this.

I assume that the rest of the answer rely's on my using the reduction formula for trigonometric integrals?

I've had a quick scoot on the web and there seems to be two main cases m+n odd or m+n even. Which uses two different formula (which does make me a little anxious as there is only one formula that we are trying to prove.)

Anyway I've included a link to the worksheet I am working from.

http://www.math.uAlberta.ca/~apotapov/MATH115/trinth.pdf

Is this the correct method to solve the problem?

 

[Bazman]

Hi VietDao,

I managed to figure it out.

Your hint brought me a lot closer to the solution than I initially realized!

Thanks again