Math 1220, Spring 2014
Instructor: Marina Gresham
Trig Sub
Steps for using a trig sub:
(1) Make substitution: (function) = (constant)TRIG2 θ
(2) Work with above substitution to find dx and anything else that appears in the
original integral in terms of θ. Then rewrite integral in terms of θ.
(3) Use trig identity to simplify integrand.
(4) Integrate.
(5) Sub back in for θ using your substitution to find θ and any trig functions appearing
in your answer (maybe use a right triangle).
Identity
What You See
Trig Function
(constant) - (function)
sinθ
1 + tan2 θ = sec2 θ (constant) + (function)
tanθ
sec2 θ
secθ
1 − sin2 θ
=
−1=
cos2 θ
tan2 θ
(function) - (constant)
Substitution: (function) = (constant)TRIG2 θ
“What you see” matches the format of the trig identity you
should use. This can help you to remember which substitution
to make.
Z
Example:
e3x
√
dx
4 − e2x
(1) Under the square root, we have (constant) - (function), where “constant” = 4
and “function” = e2x . So we’ll use sin θ and set:
e2x = 4 sin2 θ
(2) We need to be able to replace everything in the original integral in terms of θ.
First find dx:
e2x = 4 sin2 θ
p
√
2x
e = 4 sin2 θ
ex = 2 sin θ
ex dx = 2 cos θdθ
We have a copy of ex dx in our integral:
e3x
Z
Z
√
dx =
4 − e2x
e2x
√
ex dx
2x
4−e
[[If we didn’t have an extra ex in our integral, we could continue by writing:
cos θ
2 cos θ
2 cos θ
dθ
=
dθ]]
dθ
=
ex
2 sin θ
sin θ
dx =
We can rewrite our integral as:
e2x
Z
√
x
4 − e2x
4 sin2 θ
Z
e dx =
p
4 − 4 sin2 θ
2 cos θ dθ
(3) Now use the trig identity associated with our substitution (1 − sin2 θ = cos2 θ)
and simplify.
4 sin2 θ
Z
Z
p
4 − 4 sin2 θ
2 cos θ dθ = 8
4(1 − sin2 θ)
Z
=8
4(cos2 θ)
=8
Z
Z
=4
sin2 θ cos θ
p
Z
8
=
2
sin2 θ cos θ
p
dθ
dθ
sin2 θ cos θ
dθ
2 cos θ
sin2 θ dθ
sin2 θ dθ
(4) To integrate sin2 θ, use the double-angle formula: cos 2θ = 1 − 2 sin2 θ.
Z
4
Z
2
sin θ dθ = 4
1 − cos 2θ
dθ
2
Z
(1 − cos 2θ)dθ
=2
1
= 2 θ − (sin 2θ) + C
2
= 2θ − sin 2θ + C
(5) We need to write θ and sin 2θ in terms of x. We will use our substitution:
ex = 2 sin θ → sin θ =
ex
2
ex
. To find sin 2θ, use a right triangle and the double-angle
2
formula: sin 2θ = 2 sin θ cos θ.
So θ = sin−1
ex
opp
=
and use the Pythagorean
2
hyp
Theorem to find the missing side length.
Fill in sin θ =
Now we√can find any trig function of θ, including:
4 − e2x
cos θ =
2
So plugging in for θ and sin 2θ gives:
2θ − sin 2θ + C = 2θ − 2 sin θ cos θ + C
= 2 sin−1
= 2 sin−1
√
ex
4 − e2x
+C
2
2
√
ex 4 − e2x
ex
−
+C
2
2
ex
−2
2