How to do trigonometric
substitution using the method
from the tutorial
Substitutions
You need to know the following table by heart!
Integrand contains
√
2
2
√a − x
2
2
√x + a
2
x − a2
Use substitution
x = a sin θ
x = a tan θ
x = a sec θ
Differential
dx = a cos θ dθ
dx = a sec2 θ dθ
dx = a sec θ tan θ dθ
Trigonometric functions
You also need to know the following definitions of the trigonometric functions:
sin =
cos =
tan =
sec =
csc =
cot =
opposite
hypothenuse
adjacent
hypothenuse
sin
opposite
=
adjacent
cos
hypothenuse
=
adjacent
hypothenuse
=
opposite
1
adjacent
=
opposite
tan
1
sin
1
cos
cos
=
sin
Triangles
Getting the right-angle triangles is now easy.
If you are substituting x = a sin θ, then sin θ = xa . So we can label the
opposite site to θ by x and the hypothenuse by a. The last side, the adjacent,
can be found using Pythagora’s theorem (a2 + b2 = c2 ), so in this case it is
√
a2 − x2 , which is the form that the integrand contains.
The triangles for the other two substitutions can be found similarly.
1
√
a
θ
√
x
x2 + a2
θ
a2
−
x2
x = a sin θ
x
x
√
x2 − a2
θ
a
a
x = a tan θ
x = a sec θ
Substitution
Now it’s time to substitute anything involving an x with something only involving θ and constants. We can use the triangles and the definitions of the
trigonometric functions for this.
√
For example, if x = a sin θ, then we have from the triangle that a2 − x2 /a =
cos θ.
If you have a definite integral, remember to change the lower and upper
integration limits.
After doing this, simplify the integrand as far as possible.
Integration and Backsubstitution
When the integrand is simplified, integrate using the techniques from class (e.g.
trigonometric integrals, integration by parts etc).
If the integral is definite, you can now use the changed integration limits to
find the exact value of the integral.
If the integral is indefinite, you have to substitute back to get a function
in terms of x instead of θ. The triangles are really helpful in this step again.
For example, if your substitution was x = a sin θ, then we could change sec θ to
√ a
.
a2 −x2
Finished!
2
Example
√
x3
dx. Since this involves the form x2 − 16,
x2 − 16
we are going to use the substitution x = a sec θ = 4 sec θ. So the triangle is given
by:
Z
Now consider the integral
x
√
√
x2 − 16
θ
4
x = 4 sec θ
√
Reading off from this we have x2 − 16 = 4 tan θ. We also know dx =
4 sec θ tan θ dθ, so the integral becomes
Z
Z
x3
(4 sec θ)3
√
(4 sec θ tan θ) dθ
dx =
4 tan θ
x2 − 16
Z
= 64 sec4 θ dθ
Using the rules we learned for trigonometric integrals and the identity sec2 =
tan +1 we can now evaluate the integral (use the substitution u = tan θ so that
du = sec2 θ dθ).
2
Z
√
x3
dx = 64
x2 − 16
Z
Z
= 64
sec4 θ dθ
(tan2 θ + 1) sec2 θ dθ
Z
(u2 + 1) du
3
u
= 64
+u +C
3
tan3 θ
= 64
+ tan θ + C
3
= 64
√
2
We already know tan θ = x 4−16 (can also be read off from the triangle), so
that our integral finally becomes
Z
x3
√
dx =
x2 − 16
√
x2 − 16
3
3
3
+ 16
p
x2 − 16 + C