# how do you integrate cos^2(2x)?

Can I use the half angle equations to solve this? cos^2(t) = 1/2 + 1/2cos2t ?

cos^2(2x)?

I = integral

Icos^2(2x)dx= (1/2)Icos^2(t)dt

2x = t, 2dx = dt, dx = dt/2

Icos^2(t)dt= I costcostdt

u=cost, du = - sint

dv = costdt, v = sint

Icos^2(t)dt = sintcost +Isin^2(t)dt

Icos^2(t)dt =sin(2t)/2+I(1-cos^2(t))dt

Icos^2(t)dt =sin(2t)/2 + Idt - I cos^2(t)dt

Icos^2(t)dt +Icos^2(t)dt =sin(2t)/2+

2Icos^2(t)dt =sin(2t)/2 + t

Icos^2(t)dt = sin(2t)/4 + t/2

t=2x

= sin(4x)/4 + x
answered May 20, 2012 by Level 8 User (30,020 points)

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