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

Question

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

Scarlett Allison · Accepted Answer

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

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 #

Samantha Adkison · Answer

undefined To evaluate the integral ∫8cos^4(2πt) dt:

1. Use the identity cos^2(θ) = (1 + cos(2θ))/2.
2. Replace cos^4(2πt) with (1 + cos(4πt))/2.
3. Distribute the 8: 8/2 = 4.
4. Integrate term by term: ∫4(1 + cos(4πt))/2 dt.
5. Integrate each term separately: 4/2 * ∫(1 + cos(4πt)) dt.
6. Integrate each term: 4/2 * [t + (1/(4π))sin(4πt)] + C.
7. Simplify: 2t + (2/(π))sin(4πt) + C.

So, the integral of 8cos^4(2πt) dt equals 2t + (2/(π))sin(4πt) + C.