Page 222 - Calculus Demystified

CHAPTER 7
Methods of Integration
Math Note: Always be sure to check your work. You can differentiate the answer 209
in the last example to recover the integrand, confirming that the integration has
been performed correctly.
EXAMPLE 7.9
Evaluate the integral
3

2
2x x + 1 dx.
0
SOLUTION
2
We recognize that the expression 2x is the derivative of x +1. This suggests
2
the substitution u = x + 1. Thus du = 2xdx. Also x = 0 ←→ u = 1 and
x = 3 ←→ u = 10. The integral is thus transformed to
10

√
udu.
1
This new integral is a bit easier to understand if we write the square root as
a fractional power:
10 u 3/2 10 10 3/2 1 3/2 2 · 10 3/2 2

u 1/2 du = = − = − .
1 3/2 1 3/2 3/2 3 3
You Try It: Evaluate the integral

5
dx
.
3 x · log |x|
Math Note: Just as with integration by parts, we always have the option of first
evaluating the indefinite integral and then evaluating the limits at the very end. The
next example illustrates this idea.

EXAMPLE 7.10
Evaluate
π/2

cos x
dx.
sin x
π/3
SOLUTION
Since cos x is the derivative of sin x, it is natural to attempt the substitution
u = sin x. Then du = cos xdx. [Explain why it would be a bad idea to let
u = cos x.] We first treat the improper integral. We find that

cos x du
dx = = log |u|+ C.
sin x u