Integrals of the Form:

Chapter 7.3March 27, 2007

a2 −u2 ,  u2 −a2 ,  a2 +u2

Evaluate:

We can solve this integral using an “inverse substitution.”

Usually when we substitute, the new variable is a function of the old one.

With inverse substitutions, the old variable is a function of the new one!

We basically use the Substitution rule in reverse!

Letting x = g(t), we have dx = g’(t)dt and we get:

e xdx∫

f (x)dx = f g(t)( )g'(t)dt∫∫

Chapter 7.3March 27, 2007

Evaluate: e xdx∫

Given integrals of the form:

We can integrate them by thinking about area.

The graph of is the top half of a circle,

The bounds for the first integral represent 1/4 of the circle where the bounds for the second integral represent 1/2 of the circle.

We can evaluate the integrals by using the formula for the area of a circle.

But what if there are no bounds? Or if there are bounds that do NOT give us sections of a circle that are easily calculated using area?

25 −x2dx0

5

∫   or   25 −x2dx−5

5

∫

25 −x2

πr2( )

What about: ?

U-substitution does not work in this case, as we are “missing” the ‘x’ needed for the ‘du’

We could evaluate the integral if the “square root” was not there!

We will use a change of variable using an Inverse Substitution which takes advantage of some trig identities.

Specifically, forms of:

25 −x2dx∫

sin2θ + cos2θ =1

Forms of:

#1

#2 Divide by to get:

#3 Add 1 to both sides to get:

For each of these, you can replace a sum or difference with a single squared term! (Eliminating the radical!)

sin2θ + cos2θ =1

1−sin2θ =cos2θ

cos2θ

sec2θ −1=tan2θ

tan2θ +1=sec2θ

sec2θ −tan2θ =1

Back to:

25 −x2dx∫

Back to x?:

25 −x2dx∫ =252θ + cosθ sinθ( )

We can use that technique for other forms:

a2 −u2

u2 −a2

  a2 +u2

u =asinθ 1−sin2θ =cos2θ

u =asecθ sec2θ −1=tan2θ

 u =atanθ tan2θ +1=sec2θ

Let’s look at making just the substitution for now (not integrating to find the answer)

x3

16 −x2 dx∫ x4 2x2 + 4( )

32∫  dx x 3x2 −4( )

−32∫  dx

Complete the appropriate trig substitution:

x2 −4x

dx∫ 3x2 + 4( )∫32( )dx