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Roberto’s Notes on Integral Calculus
Chapter 2 : Integration methods Section 11
Integration by inverse substitution
by using the tangent function
What you need to know already: What you can learn here:
How to use inverse trig substitution to
integrate a function involving a
2 2 2
a b x or
2 2 2
b x a form.
How to use inverse trig substitution to
integrate a function involving a
2 2 2 b x a
form.
As you have probably already figured out, integrals containing a
factor of the form
2 2 2 b x a still use an inverse substitution. In fact,
this substitution uses again the identity linking secant and tangent, but
in the opposite direction.
Strategy for integrals that contain
2 2 2 b x a
For integrands involving a factor of this form:
1. Try the inverse substitution :
2 bx a tan , bdx a sec d
2. Try to compute the new integral with suitable
methods.
3. If this works, change back to the original
variable, by using other trigonometric identities
and/or the triangle model, as necessary.
And again, the same issues presented in the earlier sections
regarding domain and extended uses apply here. All you need is
practice, so here are two examples.
Example:
2
2
x
dx
x
We use the suggested substitution:
2 3tan , 3sec , tan 3
x
x dx d
This has the triangle representation shown here.
This leads to: 2 2 2 2
2 2
9 tan 3sec 9 tan sec
9 9 tan 9
x dx d d
x
(^) (^)
2 3
9 sec 1 sec d 9 sec sec d
By using earlier information about this integrand we find that this equals:
9 sec tan ln sec tan 2 2
c
Finally, we use the triangle model to get back to x :
2 2 2
2
9 ln
9 18 2 3 3
x x x x x dx c
x
^
Boy, these integrals lead to rather unexpected conclusions!
Yes: who’d have thunk, eh? Here is another one.
Example: 2 2
dx
x x
We use the inverse substitution
2 x 2 tan , dx 2sec d , represented
by this triangle model:
This changes the integral to: 2 2
2 2 2 2
2sec sec 1 sec
4 tan 4 4 tan 2 tan^ 2sec^4 tan
d d d
^
Working with this integral seems confusing, so we change it to sine and
cosine:
2
2 2
1 1 cos 1 cos
4 cos sin 4 sin
d d
This is clearer, as it suggests another substitution, this time of the usual kind:
u sin , du cos d^ ,
This produces:
2 1 2
1 cos 1 1
4 sin 4 4
d u du u c
At this point we have to go all the way back to the original variable. First
back to :
2 2
4 4 4sin
dx c c
x x u^
And finally back to x :
2
2 2
2
dx x c c x (^) x x x
x
The fact that the integral contains no trigonometric functions suggests that
perhaps a regular substitution may work as well: try to find it!
2 4 x
x
x
2 x 9
3
Computation questions:
Evaluate the integrals proposed in question 1-4 by using an inverse substitution and, if possible, also a regular substitution.
2
2 2 4
x dx x
.
3
2 2
cosh
cosh 4
x dx x
3
2 2 3 / 2
x dx
x a
2 3/
dx x
Theory questions:
- Are inverse trig substitutions used to integrate trig functions?
What questions do you have for your instructor?
