Extra Examples of Trigonometric Substitutions
Math 121
1.
Z
x2
dx
4 − x2
Recognize difference of squares under the square root
Z
(2 sin θ)2
p
2 cos θ dθ
4 − 4 sin2 θ
Use a sine sub for a2 − x2 . Remember, sub for dx
Z
4 sin2 θ
p
2 cos θ dθ
4(1 − sin2 θ)
Work the algebra to create the identity 1 − sin2 θ = cos2 θ
√
=
=
Z
=4
Z
=4
Z
=4
Z
=4
sin2 θ
2 cos θ dθ
√ √
4 cos2 θ
The identity creates the perfect square under the root
sin2 θ
2 cos θ dθ
2 cos θ
Simplify the root of the perfect square, see what cancels
sin2 θ dθ
Prepare for the EVEN power technique
1 − cos(2θ)
dθ
2
Using the half angle identity
1 − cos(2θ) dθ
Using a u-sub if needed
Z
=2
sin(2θ)
=2 θ−
+C
2
2 sin θ cos θ
=2 θ−
+C
2
Using the Double Angle Identity
= 2 (θ − sin θ cos θ) + C
Using the Trig Triangle to unwind the Trig Sub
= 2 arcsin
x
2
−
x
√
2
4 − x2
2
!!
+ C = 2 arcsin
x
2
−
!
√
x 4 − x2
+C
2
2
x = 2 sin θ
Trig. Substitute
dx = 2 cos θdθ
x
θ
√
4 − x2 2. Try the same problem in a definite integral. You can either change your limits with the trig sub
or mark them as x limits. You should choose only one method below.
Option 1: Changing Limits
√
π
3
Z π
3
(2 sin θ)2
4 sin2 θ
p
p
2 cos θ dθ
2 cos θ dθ =
2
2
π
π
4(1
−
sin
θ)
1
4
−
4
sin
θ
6
6
Z π
Z π
Z π
2
2
3
3 sin θ
3
sin θ
sin2 θ dθ
=4
2 cos θ dθ = 4
2 cos θ dθ = 4
√ √
2
π
π
π
2 cos θ
4
cos
θ
6
6
6
π
Z π
Z π
3 1 − cos(2θ)
3
sin(2θ) 3
=4
1 − cos(2θ) dθ = 2 θ −
dθ = 2
π
π
2
2
π
6
6
6
2π
π
√
√ !
sin
π sin 3
π
3
3
π
π
π
3 = 2
− −
= 2
−
− +
=
3 −
2
6
2
3
4
6
4
3
Z
3
√
x2
dx =
4 − x2
Z
Recall: x = 2 sin θ ⇒ θ = arcsin
x
2
.
1
π
x = 1 ⇒ θ = arcsin
=
2
6
Change Limits:
x=
√
3 ⇒ θ = arcsin
√ !
3
π
=
2
3
Note: If you change your limits, you do not need the Trig Triangle to return to x-variable.
Option 2: Mark Limits
√
Z
1
3
x2
√
dx =
4 − x2
Z
√
x= 3
x=1
(2 sin θ)2
p
2 cos θ dθ = · · · = 2 (θ − sin θ cos θ)
4 − 4 sin2 θ
√
x= 3
x=1
using antiderivative from above in 1.
! √
!
√
√ !
√ √
√
3
x 4 − x2
1
1 4−1
3
3 4−3
= 2 arcsin
−
= 2 arcsin
−
− 2 arcsin
−
2
2
2
2
2
2
1
π √3
π √3
π π
π
π
2
−
−2
+
=2
−
=2
=
3
2
6
2
3
6
6
3
x 3.
Z
1
5 dx
(9 + x2 ) 2
Z
1
2
=
5 3 sec θ dθ
√
9 + 9 tan2 θ
1
2
p
5 3 sec θ dθ Work the algebra to create the identity 1 + tan2 θ = sec2 θ
9(1 + tan2 θ)
Z
1
2
√ √
5 3 sec θ dθ
2
9 sec θ
=
3
= 5
3
Z
1
34
Z
1
= 4
3
Z
1
81
Z
1
=
81
Z
1
81
Z
=
=
Use a tangent sub for a2 + x2 . Remember, sub for dx
Z
=
=
Recognize sum of squares under the square root
The identity creates the perfect square under the root
1
sec2 θ dθ
sec5 θ
Simplify the root of the perfect square, see what cancels
1
dθ
sec3 θ
Flip the secant to cosine
cos3 θ dθ
Prepare for the ODD power technique
cos2 θ cos θ dθ
Isolate one copy of cosine
1 − sin2 θ cosθ dθ
1 − w2 dw
1
=
81
w3
w−
+C
3
1
=
81
sin3 θ
sin θ −
+C
3
Convert using trig identity
Finish with a substitution
1
=
81
√
u = 3 tan θ
Trig. Substitute
du = 3 sec2 θdθ
x
1
√
−
2
3
9+x
x
√
9 + x2
3 !
+C
x2 + 9
x
θ
3
w = sin θ
Standard substitution
dw = cos θ dθ Z
4.
x2
1
x arcsin x dx =
arcsin x −
2
2
Z
√
x2
dx
1 − x2
sin2 θ
x2
1
=
arcsin x −
2
2
Z
x2
1
p
· cos θ dθ =
arcsin x −
2
2
2
1 − sin θ
x2
1
=
arcsin x −
2
2
Z
sin2 θ
x2
1
· cos θ dθ =
arcsin x −
cos θ
2
2
x2
1
=
arcsin x −
2
2
Z
1 − cos(2θ)
x2
1
dθ =
arcsin x −
2
2
4
Z
Z
sin2 θ
√
· cos θ dθ
cos2 θ
sin2 θdθ
Z
1 − cos(2θ)dθ
=
x2
1
1
x2
1
1
arcsin x −
θ − sin(2θ) + C =
arcsin x − θ + sin(2θ) + C
2
4
2
2
4
8
=
x2
1
1
x2
1
1 p
arcsin x − θ + (2 sin θ cos θ) + C =
arcsin x − arcsin x + x 1 − x2 + C
2
4
8
2
4
4
u = arcsin x
IBP to start:
du = √
dv = xdx
1
x2
dx v =
2
1 − x2
1
x = sin θ
Trig. Substitute
dx = cos θdθ
x
θ
√
1 − x2
Math 121 Extra Examples of Trigonometric Substitutions
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