Integration!!!!!!!!!!!!!!!!!!!!!

Question

∫

int(cos^4x)dx

Answer 1

We have the formula (cos x)^2 = [1 + cos(2x)]/2   

Thus (cos x)^4 = [1 + cos(2x)]^2 / 4

                        = [1 + 2cos(2x) + {cos(2x)}^2] / 4

                        = [1 + 2cos(2x) + {1 + cos(4x)}/2] / 4

                        = [3 + 4cos(2x) + cos(4x)] / 8

Integrating on both sides we get,

∫ cos^4(x) dx = ∫ [3 + 4cos(2x) + cos(4x)] / 8 dx                     [ Since ∫cos(ax)dx = sin(ax)/a and ∫adx = ax ]

                     = 1/8 [3x + 2sin(2x) + 1/4 sin(4x)] + C

Therefore ∫ cos^4(x) dx = 1/8 [3x + 2sin(2x) + 1/4 sin(4x)] + C