Calculus- Early Transcendentals, 2021a

254 Techniques of Integration
u, thus getting the incorrect answer − 1 2 cos(4)+1 cos(2). An acceptable alternative is something like:
2
∫ 4
2
xsin(x 2 )dx =
∫ x=4
x=2
∣
1
∣∣∣
x=4
2 sinudu= −1 2 cos(u)
x=2
= − 1 2 cos(x2 )
∣
4
2
= − cos(16)
2
+ cos(4) .
2
♣
To summarize, we have the following.
Theorem 7.9: Substitution Rule for Definite Integrals
If g ′ is continuous on [a,b] and f is continuous on the range of u = g(x),then
∫ b
a
f (g(x))g ′ (x)dx =
∫ g(b)
g(a)
f (u)du.
Example 7.10: Substitution Rule
∫ 1/2 cos(πt)
Evaluate
1/4 sin 2 (πt) dt.
Solution. Let u = sin(πt) so du = π cos(πt)dt or du/π = cos(πt)dt. We change the limits to sin(π/4)=
√
2/2andsin(π/2)=1. Then
∫ 1/2
1/4
∫
cos(πt) 1
∫
sin 2 (πt) dt = 1 1 1
√
2/2 π u 2 du =
√
2/2
1
π u−2 du = 1 u −1
1
π −1 ∣√ = − 1 √
2
2/2
π + π .
♣
Exercises for Section 7.1
Find the following indefinite and definite integrals.
∫
Exercise 7.1.1 (1 −t) 9 dt
Exercise 7.1.2
Exercise 7.1.3
Exercise 7.1.4
∫
∫
∫
(x 2 + 1) 2 dx
x(x 2 + 1) 100 dx
1
3√
1 − 5t
dt