Page 221 - Calculus Demystified

CHAPTER 7
Methods of Integration
208 To see a model situation, imagine an integral
b
f(x) dx.
a
If the techniques that we know will not suffice to evaluate the integral, then we might
attempt to transform this to another integral by a change of variable x = ϕ(t). This

entails dx = ϕ (t) dt. Also
x = a ←→ t = ϕ −1 (a) and x = b ←→ t = ϕ −1 (b).
Thus the original integral is transformed to

ϕ −1 (b)
ϕ −1 (a) f(ϕ(t)) · ϕ (t) dt.

It turns out that, with a little notation, we can make this process both convenient
and straightforward.
We now illustrate this new paradigm with some examples. We begin with an
indefinite integral.

EXAMPLE 7.8
Evaluate TEAMFLY
5
[sin x] · cos xdx.

SOLUTION
On looking at the integral, we see that the expression cos x is the derivative
of sin x. This observation suggests the substitution sin x = u. Thus cos xdx =
du. We must now substitute these expressions into the integral, replacing all
x-expressions with u-expressions. When we are through with this process, no
x expressions can remain. The result is

5
u du.

This is of course an easy integral for us. So we have
5 5 u 6
[sin x] · cos xdx = u du = 6 + C.
Now the important final step is to resubstitute the x-expressions in place of
the u-expressions. The result is then

6
[sin x] · cos xdx = sin x + C.
5
6

®
Team-Fly