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

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?

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

Thank you very much!

Your approach is correct in using the binomial expansion to expand (sinx+cosx)^2n. This will allow you to integrate the expression and solve for I.

To simplify the process, you can use the concept of symmetry. Notice that the integral you are trying to solve, ∫[(sinx)^2n dx] from 0 to π/2, involves the sin function. However, the original expression is symmetric about x = π/4. Therefore, you can rewrite the original integral as:

∫[(sinx)^2n dx] from 0 to π/2 = 2 * ∫[(sinx)^2n dx] from 0 to π/4

Now, you can apply your method of expanding (sinx+cosx)^2n using the binomial coefficients. The sum of the terms with odd powers of cosx will evaluate to zero due to the symmetric bounds of integration. Therefore, you only need to consider the terms with even powers of cosx.

Once you have expanded (sinx+cosx)^2n and kept only the even powers, integrate each term individually. You will notice that the integration of each term involves expressions like ∫[(cosx)^i dx] and ∫[(sinx)^(2n-i) dx], which are known formulas. Apply these formulas to simplify the integrals.

Ultimately, you should obtain an expression for the integral in terms of factorials and powers of π. Simplify the expression and compare it to [(2n)!*π]/[(2)^(2n+1)*(n!)^2]. If they are equal, you have successfully proven the given identity.

Note: This process can involve a lot of algebraic manipulation and working with factorials. Make sure to pay attention to indices and signs while performing the calculations.