ABSTRACT
Problem 1. Evaluate ∫ π
0 2 cos2 4t dt.
Since cos 2t = 2 cos2 t − 1 (from Chapter 18),
then cos2 t = 1 2
(1 + cos 2t) and
cos2 4t = 1 2
(1 + cos 8t)
Hence ∫ π
0 2 cos2 4t dt
= 2 ∫ π
1 2
(1 + cos 8t) dt
= [
t + sin 8t 8
= ⎡
⎢ ⎣
π
4 +
sin 8 (π
)
⎤
⎥ ⎦ −
[
0 + sin 0 8
]
= π 4
or 0.7854
Problem 2. Determine ∫
sin2 3x dx.
