# integral of trig function

• Apr 27th 2007, 10:00 PM
integral 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, 01:23 AM
ecMathGeek
Quote:

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

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