This tutorial assumes that you are familiar with trigonometric identities, derivatives, integration of trigonometric functions, and integration by substitution.
After simpler methods of integration failed, we should consider trigonometric substitution. The requirement is that the function contains the form
a2−x2
, the form
a2+x2
, or the form
x2−a2
. Some examples are:
1.
x2−25
2.
(4−y2)23
3.
9−h2
4.
16+x21
In fact, there are clear rules on what substitutions to make:
1. If the function contains
a2−x2
, let
x=asinu
.
2. If the function contains
a2+x2
, let
x=atanu
.
3. If the function contains
x2−a2
, let
x=asecu
.
Using Trigonometric Substitution
Let's try an example. Solve this integral:
∫16+x21dx
Since the denominator fits the form
a2+x2
, let's use Rule #2 from the section above.
We let
x=atanu
, which is
4tanu
.
Then, to get
dx
, we find the derivative of
4tanu
, which is
4sec2u
.
Therefore, let
x=atanu
, and
dx=4sec2u
.
Substituting the variables from the above, we have:
∫16+(4tanu)21×4sec2udu
After simplifying, we have:
∫41du
By the Power Rule, the integral is:
4u
From the earlier steps, we know that
x=atanu
, and solving for
u
, we have:
u=tan−1(41x)
The only step remaining is to substitute the value of
u
back into the earlier integral, namely
4u
. After the substitution, we have:
4tan−1(41x)+C
We are done. We have successfully used trigonometric substitution to find the integral.
What's Next
Ready to dive deeper? You can try more practice problems at the top of this page to help you get more familiar with solving integral using trigonometric substitution. Want even more help? Sign up for Cymath Plus today. You can also download the Cymath app for iOS and Android to get step-by-step help on-the-go.