Chapter 5 | Integration
591
Using Substitution to Evaluate a Trigonometric Integral
Use substitution to evaluate ∫ 0 π /2
cos 2 θdθ .
Solution Let us first use a trigonometric identity to rewrite the integral. The trig identity cos 2 θ = 1+cos2 θ 2
allows us
to rewrite the integral as
π /2 1+cos2 θ 2
π /2
cos 2 θdθ = ⌠ ⌡ 0
∫
dθ .
0
Then,
π /2 ⎛
⎞ ⎠ dθ = ⌠
π /2 ⎛
θ ⎞
⌠ ⌡ 0
⎝ 1+cos2 θ 2
⎝ 1
1 2 cos2
⎠ dθ
⌡ 0
2 +
π /2
π /2
= 1 2 ∫
2 ∫
dθ + 1
cos2 θdθ .
0
0
We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let u =2 θ . Then, du =2 dθ , or 1 2 du = dθ . Also, when θ =0, u =0, andwhen θ = π /2, u = π . Expressing the second integral in terms of u , we have
π /2
⌠ ⌡ 0
π /2
⌠ ⌡ 0
⎛ ⎝ 1 2
⎞ ⎠ ∫
π /2
π
1 2
2 ∫
dθ + 1
cos2 θdθ = 1 2
dθ + 1 2
cos udu
0
0
2 | θ =0 4 −0 ⎛ ⎝ π
u |
θ = π /2
u =0 u = θ
= θ
+ 1 4 sin
⎞ ⎠ +(0−0) = π 4 .
=
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