How do you find the definite integral for: #sin^(2) (5x) * cos^(2) (5x) dx # for the intervals #[8, 11]#?

1 Answer
Apr 7, 2018

#~~0.3758#

Explanation:

We use the double angle formula for the sine function

#sin(2A) = 2 sin A cos A#

to write

# sin^(2) (5x) * cos^(2) (5x) = (sin(5x) * cos (5x))^2 = (1/2 sin (10x))^2 = 1/4 sin^2(10x)#

We make use of the double angle formula for the cosine

#cos(2A) = cos^2 A-sin^2 A = 1-2sin^2 A#

to write the integrand in the form

#1/4 (1/2[1-cos(20x)]) = 1/8(1-cos(20x))#

So, finally

#int_8^11 sin^(2) (5x) * cos^(2) (5x) dx = int_8^11 1/8(1-cos(20x)) dx#
#qquad = 1/8 (x-1/20 sin(20x))_8^11 #
#qquad = 1/8((11-8)-(sin(220)-sin(160))/20) ~~ 0.3758#