Math (Integrals) (Using given method)

I would like to solve the ∫sin^2(pix) dx

Using the given substitution identity: sin^2(x) = (1/2)(1-cos2x)

This is what I did so far:

∫sin^2(pix) let u = pix
du = pi dx

(1/pi)∫sin^2(u)du

Applying the identity is where I'm lost on how to continue further..

(1/pi)∫(1/2)(1-cos2u?) du or is it cos2x?

I went further with cos2u by doing:

(1/2pi)∫(1-cos2u) du
(1/2pi)(x - (sin(2pix)/2pi)) +C

However, I'm pretty sure I messed up on the cos2x part of substituting since that answer doesn't seem correct..

Any help is greatly appreciated!

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asked by Ray
  1. ∫sin^2(π x) dx
    = ∫(1/2)(1-cos(2πx) ) dx ---> you had that
    = ∫(1/2 - (1/2)(cos 2x) dx

    At this point you should be able to put your brain in reverse and ask yourself, "what do I need so that the derivative is 1/2 - (1/2)(cos2πx)

    remembering that d(sin kx)/dx = k cos kx
    I would get
    (1/2)x - (1/4π )sin (2πx) + c

    you could even get fancy by recalling that
    sin (2πx) = 2(sin πx)(cos πx) , but .....

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    posted by Reiny
  2. Thanks! So I was on the right track, but messed up near the ending.

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    posted by Ray

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