chapter  40
Integration using trigonometric and hyperbolic substitutions
Pages 11

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.