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[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

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# Homework Help: Integration by Parts

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