Hey All,

I wanted to know when you integrate sin^4x where the 4 comes from in the integral.

∫sin^4x dx= ∫(sin^2x)^2 dx= ∫(1-cos2x)^2/4 dx

thank you

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- Apr 27th 2007, 09:00 PMdadonintegral of trig function
Hey All,

I wanted to know when you integrate sin^4x where the 4 comes from in the integral.

∫sin^4x dx= ∫(sin^2x)^2 dx= ∫(1-cos2x)^2/4 dx

thank you - Apr 28th 2007, 12:23 AMecMathGeek
I'm sure you've seen the trig identities enough to be familiar with the double angle formulas:

cos(2x) = 2cos^2(x) - 1 = 1 - 2sin^2(x) = cos^2(x) - sin^2(x)

Well, for this problem, we are especially interested in the version containing only sin^2(x)

cos(2x) = 1 - 2sin^2(x)

Let's solve this for sin^2(x):

sin^2(x) = 1/2(1 - cos(2x))

Using this we can replace a sin^2(x) term (which we cannot integrate) with 1/2(1 - cos(2x)) (which we can integrate).

Thus the integration becomes:

∫sin^4x dx= ∫(sin^2x)^2 dx= ∫[1/2(1 - cos(2x))]^2 dx = ∫(1/2)^2*(1 - cos(2x)^2 dx = ∫(1-cos2x)^2/4 dx