Evaluate the integral? : #int 8cos^4 2pit dt #

1 Answer
Sep 8, 2017

# int \ 8cos^4 2pit \ dt = 3x + (sin(4pix))/(pi)+(sin(8pix))/(8pi) + C #

Explanation:

We seek:

# I = int \ 8cos^4 2pit \ dt #

We can utilise the double angle identity:

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

From which we get:

# cos^2A -= 1/2(1+cos2A) #

And squaring we get:

# cos^4A -= 1/4(1+cos2A)^2 #
# \ \ \ \ \ \ \ \ \ \ = 1/4(1+2cos2A+cos^2 2A) #
# \ \ \ \ \ \ \ \ \ \ = 1/4(1+2cos2A+1/2(1+cos4A)) #
# \ \ \ \ \ \ \ \ \ \ = 1/8(2+4cos2A+(1+cos4A)) #
# \ \ \ \ \ \ \ \ \ \ = 1/8(3+4cos2A+cos4A) #

To make thing easier to read we can also perform a simple substitution:

Let # u =2pit => (du)/dt = 2pi #

So our integral becomes:

# I = 8 int \ (cos^4 u) \ 1/(2pi) \ du #
# \ \ = 8/(2pi) int \ (cos^4 u) du #
# \ \ = 8/(2pi) int \ (1/8(3+4cos2u+cos4u)) \ du #
# \ \ = 1/(2pi) int \ 3+4cos2u+cos4u \ du #
# \ \ = 1/(2pi) {3u + 4/2sin2u+1/4sin4u } + C#
# \ \ = 3u/(2pi) + (2sin2u)/(2pi)+(1/4sin4u)/(2pi) + C#

And restraining the substitution we have:

# I = 3(2pix)/(2pi) + (2sin(4pix))/(2pi)+(1/4sin(8pix))/(2pi) + C#
# \ \ = 3x + (sin(4pix))/(pi)+(sin(8pix))/(8pi) + C #