Question

Determine the following integral for x = -
π/3 to x = π/3:
S (cos^2x-cos^4x) dx​

Answer 1

Answer:

put x = π/3

cos^2x-cos^4x=cos^2π\3-cos^4π\3

=(1/2)^2-(1/2)^4

=1/4-1/16

=(16-4)/64

=12/64.

cos^2x-cos^4x=3/16

put x = -π/3

cos^2x-cos^4x=cos^2(-π\3)-cos^4(-π\3)

=(-1/2)^2-(-1/2)^4

=1/4-1/16

=(16-4)/64

=12/64.

cos^2x-cos^4x=3/16