Trig Substitution Solutions
Written by Victoria Kala
[email protected]
SH 6432u Office Hours: R 11:00 am - 12:00 pm
Last updated 2/13/2015
Evaluate the following integrals.
Z p
1.
x 1 − x4 dx (Hint: first use the substitution u = x2 )
Solution. Let’s move some stuff around inside the integral:
Z p
Z p
Z p
4
2
2
x 1 − x dx = x 1 − (x ) dx =
1 − (x2 )2 xdx
Now let’s try the substitution u = x2 , then du = 2xdx:
Z p
Z
1 p
1 − (x2 )2 xdx =
1 − u2 du
2
We now need to use a trig substitution. Use u = sin θ, du = cos θdθ:
Z
Z
1 p
1 p
1 − u2 du =
1 − sin2 θ cos θdθ
2
2
Use the identity sin2 θ + cos2 θ = 1 to simplify underneath the square root:
Z
Z
Z
Z
1 p
1 √ 2
1
1
2
1 − sin θ cos θdθ =
cos θ cos θdθ =
cos θ cos θdθ =
cos2 θdθ
2
2
2
2
Use the half-angle identity cos2 θ = 12 (1 + cos 2θ):
Z
Z
1
1
1
1
1 1
1
·
cos2 θdθ =
(1 + cos 2θ) dθ =
θ + sin 2θ + C = θ + sin 2θ + C
2 2
4
4
2
4
8
Before we use our triangle to plug back in u, we need to simplify a bit further. Use the identity
sin 2θ = 2 sin θ cos θ:
1
1
1
1
1
1
θ + sin 2θ + C = θ + · 2 sin θ cos θ + C = θ + sin θ cos θ + C
4
8
4
8
4
4
Now let’s draw the triangle. Since u = sin θ, the opposite
side of our triangle will be u, the
√
hypotenuse will be 1, and the adjacent side will be 1 − u2 . Also note that u = sin θ implies
θ = sin−1 (u). Now plug these into our solution:
1
1
1
1 p
θ + sin θ cos θ + C = sin−1 (u) + u 1 − u2 + C
4
4
4
4
The last thing to do is plug in u = x2 :
1 p
1
sin−1 x2 + x2 1 − x4 + C
4
4
1
Z
2.
√
t2
dt
(Hint: complete the square on the denominator)
− 6t + 13
Solution. We need to complete the square of t2 − 6t + 13. Take half of the middle term (the
middle term is −6) and square it. This will be the number we plus/minus by:
(t2 − 6t) + 13 = (t2 − 6t + 9 − 9) + 13
= (t2 − 6t + 9) − 9 + 13
= (t2 − 6t + 9) + 4
= (t − 3)2 + 4
Let’s put that in our integral:
Z
√
dt
=
2
t − 6t + 13
Z
dt
p
(t − 3)2 + 4
Use the substitution u = t − 3, du = dt:
Z
Z
du
dt
p
√
=
2
u2 + 4
(t − 3) + 4
Use the trig sub u = 2 tan θ, du = 2 sec2 θdθ:
Z
Z
Z
du
2 sec2 θdθ
2 sec2 θdθ
√
√
p
=
=
u2 + 4
4 tan2 θ + 4
4(tan2 θ + 1)
Use the identity tan2 θ + 1 = sec2 θ to simplify underneath the square root:
Z
Z
Z
Z
2 sec2 θdθ
2 sec2 θdθ
2 sec2 θdθ
p
√
=
=
sec θdθ = ln | sec θ + tan θ| + C
=
2 sec θ
4 sec2 θ
4(tan2 θ + 1)
Let’s draw the triangle. Since u = 2 tan θ, then u2 = tan θ. The opposite side of our triangle
√
will be u, the adjacent side will be 2, and the hypotenuse will be u2 + 4. Plug these into
our solution:
√
u2 + 4 u + +C
ln | sec θ + tan θ| + C = ln 2
2
Then plug back in u = t − 3:
p
(t − 3)2 + 4 t − 3 ln +
+C
2
2 Z √
3.
x2 − 4
dx
x
2
Solution. Use the trig sub x = 2 sec θ, dx = 2 sec θ tan θdθ:
Z √ 2
Z √
Z p
x −4
4 sec2 θ − 4
4(sec2 θ − 1) tan θdθ
dx =
2 sec θ tan θdθ =
x
2 sec θ
Use the identity tan2 θ + 1 = sec2 θ to simplify underneath the square root:
Z p
Z p
Z
Z
4(sec2 θ − 1) tan θdθ =
4 tan2 θ tan θdθ = 2 tan θ tan θdθ = 2 tan2 θdθ
Use the identity tan2 θ + 1 = sec2 θ again:
Z
Z
2 tan2 θdθ = 2
sec2 θ − 1 dθ = 2 (tan θ − θ) + C
Let’s draw the triangle. Since x = 2 sec θ, then x2 = sec θ. The hypotenuse is x, the adjacent
√
side is 2, and the opposite side is x2 − 4. Note that θ = sec−1 ( x2 ). Plug these into our
solution:
!
√
x
p
x2 − 4
−1 x
2 (tan θ − θ) + C = 2
− sec
+ C = x2 − 4 − 2 sec−1
+C
2
2
2
3