Question: For any positive integer n,show that integrate.[(sin x)^2n dx ] from o - π/2 = [(2n)!*π]/[(2)^(2n+1)*(n!)^2 ]

Question

What I thought:
Let I =int.[(sinx)^2n dx]
And again I= int.[ (sin(π/2-x))^2n dx] = int.[ (cos)^2n dx]
2I= int.[(sin x)^2n + (cos x)^2n dx]
2I= int.[ (sin x+cos x)^2n - [sigma { 2nCr*(sin x)^(2n-r)*(cosx)^r r=1 --> r=(2n-1) } ] dx]

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{Let T(r+1) be the (r+1)th term of the expansion (sinx+cosx)^2n

T(r+1)= 2nCr*(sin x)^(2n-r)*(cos x)^r ( 0<=r<=(2n+1) }

Is this the correct method? Or is there any other cleverer and simple method?

Answer 1

Hmmm. That's a possibility. You might also try doing integration by parts twice to yield the recursive sequence derived here:

https://math.stackexchange.com/questions/416162/a-recursive-formula-for-a-n-int-0-pi-2-sin2nxdx-namely-a-n

Answer 2

Thank you very much!