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Practice Problems: Trig Substitution
Written by Victoria Kala
vtkala@math.ucsb.edu
November 9, 2014
The following are solutions to the Trig Substitution practice problems posted on November 9.
R √ 1 −1
1. Use trig substitution to show that 2 dx = sin x+C
1−x
Solution: Let x = sinθ, then dx = cosθ:
Z √ 1 Z p 1 Z cosθ Z −1
2dx = 2 · cosθdθ = cosθdθ = dθ = θ +C = sin x+C
1−x 1−sin θ
R 1 −1
2. Use trig substitution to show that 1+x2dx = tan x+C
2
Solution: Let x = tanθ, then dx = sec xdx:
Z 1 Z 1 2 Z sec2θ Z −1
1+x2dx= 1+tan2θ ·sec θdθ = sec2 θdθ = dθ = θ +C = tan x+C
3. Z √1 dx
2 2
x x +4
2
Solution: Let x = 2tanθ, then dx = 2sec θdθ:
Z Z Z 2
√1 √1 2 2sec θ
2 2 dx = 2 2 · 2 sec θdθ = 2 dθ
x x +4 4tan θ 4tan θ+4 4tan θ·2secθ
=Z secθ dθ= 1Z cosθdθ
2 4 2
4tan θ sin θ
Now let w = sinθ, then dw = cosθdθ:
1 Z cosθ dθ = 1 Z 1 dw = − 1 +C =−1cscθ+C
4 sin2 θ 4 w2 4w 4
Next, we need to plug back in x. Originally we had the substitution x = 2tanθ, so tanθ = x.
√ 2
This means our opposite side is x, our adjacent side is 2, and the hypotenuse is x2 +4.
Then we have Z √
1 1 x2 +4
2√ 2 dx = −4 cscθ+C =− 4x +C
x x +4
1
Z √ 2
4. 1+x dx
x
2
Solution: Let x = tanθ, then dx = sec θdθ:
Z √1+x2 Z √1+tan2θ Z secθsec2θ Z secθ(1+tan2θ)
dx = · sec2 θdθ = dθ = dθ
x tanθ tanθ tanθ
Z 2 Z
= secθ + secθtan θ dθ = (cscθ +secθtanθ)dθ = ln|cscθ −cotθ|+secθ+C
tanθ tanθ
Our substitution was x = tanθ, so our triangle has opposite side x, adjacent side 1, and
√ 2
hypotenuse x +1. Then
Z √ √
2 2 p
1+x x +1 1 2
dx = ln|cscθ −cotθ|+secθ+C = ln − + x +1+C
x x x
Ra 2√ 2 2
5. 0 x a −x dx
Solution: Let x = asinθ, then dx = acosθdθ. We should also change the bounds. When
x=a,sinθ =1⇒θ= π. When x=0,sinθ=0⇒θ=0.
2
Z a 2p 2 2 Z π/2 2 2 p 2 2 2 4 Z π/2 2 2
0 x a −x dx= 0 a sin θ a −a sin θ·acosθdθ =a 0 sin θcos θdθ
Use half angle identity:
Z π/2 Z π/2 4 Z π/2
4 2 2 4 1 1 a 2
a 0 sin θcos θdθ = a 0 2 (1 −cos2θ) 2 (1+cos2θ)dθ = 4 0 (1 −cos 2θ)dθ
4 Z π/2 4 Z π/2 4 4
a a 1 a 1 π/2 a π
2 2
= 4 sin 2θdθ = 4 2(1−cos 4θ)dθ = 8 θ − 4 sin4θ = 16
0 0 0
6. Z 0.6 √ x2 2dx
0 9−25x
Solution: Let’s simplify the integral a little bit:
Z 0.6 2 Z 0.6 2 Z 0.6 2
√ x dx = q x dx = q x dx
0 9−25x2 0 9(1− 25x2) 0 3 1−(5x)2
9 3
Let 5x = sinθ, then x = 3 sinθ and dx = 3 cosθdθ. We should also change the bounds.
3 5 5
When x=0.6,sinθ =1⇒θ= pi. When x=0,sinθ =0⇒θ =0. Then
2
Z 0.6 2 Z π/2 (3 sinθ)2 Z π/2 33sin2θcosθ
x 5 3 5
0 q 5 2dx = 0 p 2 · 5 cosθdθ = 0 3cosθ dθ
3 1−(3x) 3 1−sin θ
2
9 Z π/2 9 1 Z π/2 9 1 π/2 9π
2
= 125 0 sin θdθ = 125 · 2 0 (1 −cos2θ)dθ = 250 θ − 2 sin2θ 0 = 500
R1√ 2
7. 0 x +1dx
Solution: Let x = tanθ, then dx = sec2θdθ. When x = 1,tanθ = 1 ⇒ θ = π. When
4
x=0,tanθ=0⇒θ=0. Then
Z 1p 2 Z π/4p 2 2 Z π/4 3
0 x +1dx= 0 tan θ+1·sec θdθ = 0 sec θdθ
Wedid this integral on a previous practice sheet, you just need to use integration by parts:
Z π/4
1 π/4 1 √ √
3
sec θdθ = (secθtanθ+ln|secθ+tanθ|) = 2+ln( 2+1)
2 2
0 0
8. Z √ 2 x dx
x +x+1
Solution: Complete the square: x2 +x+1 = (x+ 1)2 + 3. Then
Z Z 2 4
√ x dx = q x dx
x2 +x+1 (x+ 1)2 + 3
2 4
Let u = x+ 1, then du = dx:
2
Z x Z u−1 Z u 1 Z 1
q dx = q 2 du= q du− q du
1 2 3 2 3 2 3 2 2 3
(x+ 2) + 4 u + 4 u + 4 u + 4
√ √
Wewill use the same substitution for both integrals. Let u = 3 tanθ, then du = 3 sec2 θdθ:
2 2
Z Z Z √3 tanθ √ Z √
q u du−1 q 1 du = q 2 3 sec2 θdθ−1 q 1 3 sec2 θdθ
2 3 2 2 3 3 2 3 2 2 3 2 3 2
u + 4 u + 4 4 tan θ + 4 4 tan θ + 4
Z √ √ Z √ √ Z Z
3 tanθ 3 sec2θ 1 3 sec2 θ 3 1
2 2 2
= √ dθ − √ dθ = secθtanθdθ− secθdθ
3 secθ 2 3 secθ 2 2
2 √ 2
= 3secθ−1ln|secθ+tanθ|+C
2 2 √
Nowweneedtoplug u back in. Our substitution was u = 3 tanθ, so our triangle with have
√ q 2
3 2 3
u on the opposite side, 2 on the adjacent side, and u + 4 on the hypotenuse. So then
√ r q
2 3
3 1 3 1 2 u + 4 2u
2 √ √
2 secθ− 2 ln|secθ+tanθ|+C = u + 4 − 2 ln 3 + 3 +C
3
Next plug back in x:
s q
Z 1 2 3
x 1 2 3 1 2 (x+ ) + 2(x+ 1)
q dx = x+ + − ln √2 4 + √ 2 +C
1 2 3 2 4 2 3 3
(x+ 2) + 4
9. Z x2 2 3/2dx
(3+4x−4x )
2 2 2 1 1
Solution: Complete the square: 3 + 4x − 4x = 3 − 4(x − x) = 3 − 4(x − x + 4 − 4) =
3−4(x−1)2+1=4−4(x−1)2. So then
2 2
Z 2 Z 2 Z 2
x dx = x dx = x dx
q q
2 3/2 3 3
(3+4x−4x ) 4−4(x−1)2 8 1−(x−1)2
2 2
Let u = x− 1, du = dx:
2
Z 2 Z (u+ 1)2 Z u2+u+1
x dx = 2 dx = 1 √ 4 du
q √
3 2 3 8 2 3
1 2 8 1−u ( 1−u )
8 1−(x−2)
Z 2 Z Z
= 1 √ u du+ √ u du+1 √ 1 du
8 2 3 2 3 4 2 3
( 1−u ) ( 1−u ) ( 1−u )
Now let u = sinθ, so du = cosθdθ:
!
= 1 Z p sin2θ2 3 cosθdθ+Z p sinθ2 3 cosθdθ+ 1Z p 1 2 3 cosθdθ
8 ( 1−sin θ) ( 1−sin θ) 4 ( 1−sin θ)
Z 2 Z Z
= 1 sin θcosθdθ+ sinθcosθdθ+ 1 cosθ dθ
3 3 3
8 cos θ cos θ 4 cos θ
= 1 Z sin2θdθ+Z sinθ dθ+ 1Z 1 dθ
8 cos2θ cos2θ 4 cos2θ
1 Z 2 Z 1 Z 2
= 8 tan θdθ+ secθtanθdθ+ 4 sec θdθ
= 1 Z (sec2θ−1)dθ+Z secθtanθdθ+ 1Z sec2θdθ
8 4
= 1 tanθ−θ+secθ+ 1tanθ+C = 15tanθ−θ+secθ+C
8 4 8 4
Now we need to plug u back in. Our substitution was u = sinθ, so the opposite side will be
√ 2
u, the hypotenuse will be 1, and the adjacent side will be 1−u :
1 5 1 5√ u −1 √ 1
8 4 tanθ −θ+secθ +C = 8 4 2 −sin u+ 2 +C
1−u 1−u
4
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