how do you integrate sin^4(x)????

Answer 1

You look it up in a Table of Integrals, like I just did:

A = bx
In your case b = 1

indefinite integral of sin^4(A) =
-(sin(2A))/4b + (sin(4A)/32b) + 3x/8

.

Answer 2

sin(x)^4 = sin(x)^2*(1-cos(x)^2) = sin(x)^2 - (sin(x)*cos(x))^2 = sin(x)^2 - 1/4*(2*sin(x)*cos(x))^2

= sin(x)^2 - 1/4*sin(2x)^2

= 1/2*(1 - cos(2x)) - 1/4*1/2*(1 - cos(4x))

integrate

= 1/2*(x - sin(2x)/2) - 1/4*1/2*(x - sin(4x)/4)
= 3x/8 - sin(2x)/4 + sin(4x)/32

Answer 3

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

You do something like that and integrate.

EDIT: NO!
Do (sin^2(x))^2
INT(1/2 - cos2x/2)^2dx
1/4INT(1-2cos2x +cos^2x)dx
1/4int[1-2cos2x + 1/2(1+cos4x)]dx
=1/4int[3/2 - 2cos2x +1/2cos4x]dx
= 1/4(3/2x - sin2x + 1/8sin4x) + C