Page 171 - Calculus Demystified

158 EXAMPLE 6.11 CHAPTER 6 Transcendental Functions
Evaluate the integral
3 2
[exp(cos x)]· sin x · cos xdx.
SOLUTION
For clarity, we let ϕ(x) = cos x, ϕ (x) = 3 cos x · (− sin x). Then the
3
2

integral becomes
1 1
− 3 exp(ϕ(x)) · ϕ (x) dx =− 3 exp(ϕ(x)) + C.

Resubstituting the expression for ϕ(x) then gives

[exp(cos x)]· sin x · cos xdx =− 1 exp(cos x) + C.
3
3
2
3
EXAMPLE 6.12
Evaluate the integral

exp(x) + exp(−x) dx.
exp(x) − exp(−x)
SOLUTION TEAMFLY
For clarity, we set ϕ(x) = exp(x) − exp(−x), ϕ (x) = exp(x) + exp(−x).

Then our integral becomes
ϕ (x) dx

ϕ(x) = ln |ϕ(x)|+ C.
Resubstituting the expression for ϕ(x) gives

exp(x) + exp(−x) dx = ln | exp(x) − exp(−x)|+ C.
exp(x) − exp(−x)
You Try It: Calculate x · exp(x − 3)dx.
2

6.2.3 THE NUMBER e
The number exp(1) is a special constant which arises in many mathematical and
physical contexts. It is denoted by the symbol e in honor of the Swiss mathematician
Leonhard Euler (1707–1783) who first studied this constant. We next see how to
calculate the decimal expansion for the number e.
In fact, as can be proved in a more advanced course, Euler’s constant e satisfies
the identity
1 n
n→+∞ 1 + n = e.
lim

Team-Fly
®