How do you evaluate #int sin^2(5x)cos^2(5x) dx# for [7, 11]?

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Answer 1 · 2015-09-16T08:30:19

#1/2+1/80sin40#

Lets transform integrand using formulas:
#sin^2(x/2)=(1-cosx)/2, cos^2(x/2)=(1+cosx)/2#

In our case:

#sin^2(5x)cos^2(5x)=(1-cos10x)/2 (1+cos10x)/2=#

#=(1-cos^2(10x))/4=1/4-1/4cos^2(10x)=#

#=1/4-1/4*(1+cos20x)/2=1/4-1/8-1/8cos20x=#

#=1/8-1/8cos20x#

We have (using substitution #t=20x, dt=20dx, dx=dt/20# for #cos20x# part):

#int(1/8-1/8cos20x)dx = 1/8x-1/8*1/20sin20x+C=#
#=1/8x-1/160sin20x+C=I#

Changing the values we get:

#I=11/8-1/160sin220-7/8+1/160sin40=#
#=1/2-1/160(sin220-sin140)=#
#=1/2-1/160*2cos(360/2)sin(80/2)=#
#=1/2-1/80(-1)sin40=1/2+1/80sin40#