Download 4.4 Trigonometric Substitutions

4.4 Trigonometric Substitutions
4.4
Brian E. Veitch
Trigonometric Substitutions
Trig substitution reduces integrals of a certain form to integrals of trig functions, which can
be easier to deal with. The idea is to match the integral like so
If you see
√
a2 − x 2
√
a2 + x 2
√
x 2 − a2
Substitute
Uses the following Identity
x = a sin(θ)
1 − sin2 (θ) = cos2 (θ)
x = a tan(θ)
1 + tan2 (θ) = sec2 (θ)
x = a sec(θ)
sec2 (θ) − 1 = tan2 (θ)
I think it’s best we just get right into it.
Z
Example 4.5. Find
√
dx
25 + x2
Since the bottom has the form a2 + x2 , we use the second option.
Let x = 5 tan(θ), dx = 5 sec2 (θ) dθ. So,
Z
dx
√
=
25 + x2
Z
Z
5 sec2 (θ) dθ
p
=
25 + 25 tan2 (θ)
5 sec2 (θ) dθ
=
5 sec(θ)
5 sec2 (θ) dθ
Z
p
25(1 + tan2 (θ))
Z
=
5 sec2 (θ) dθ
p
25 sec2 (θ)
Z
sec(θ) dθ = ln | sec(θ) + tan(θ)| + C
That didn’t seem to bad, right? Well, we’re not done yet. Remember, this is a substitution problem. The original variable is x. So we need to substitute back. Sometimes it’s
easy, and sometimes you need a little trig.
137
4.4 Trigonometric Substitutions
Brian E. Veitch
1. So what’s tan(θ)
Since x = 5 tan(θ), we get tan(θ) =
x
.
5
2. And sec(θ) ?
Let’s take a look at a right triangle with angle θ.
√
25 + x2
.
From here, we see that sec(θ) =
5
Therefore,
Z
or
Z
√
25 + x2 x dx
√
= ln + +C
2
5
5
25 + x
√
dx
√
= ln | 25 + x2 + x| − ln 5 + C
25 + x2
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4.4 Trigonometric Substitutions
Z
Example 4.6. How about
√
Brian E. Veitch
x dx
25 + x2
You need to be careful. When you’re working through a textbook, all the problems in
those sections are usually done using the same method. In this case that method would be
trig substitution. And it would work. But, this integral can be done using u-substitution.
1. Using Trig Substitution
Let x = 5 tan(θ), dx = 5 sec2 (θ) dθ
Z
x dx
√
=
25 + x2
Z
25 tan(θ) sec2 (θ) dθ
p
=
25 sec2 (θ)
Z
25 tan(θ) sec2 (θ) dθ
=
5 sec(θ)
Z
√
√
25 + x2
5 sec(θ) + C = 5
+ C = 25 + x2 + C
5
We used the right triangle from above to change the sec(θ).
2. Using u-substitution
Let u = 25 + x2 , du = 2x dx →
Z
du
= x dx
5
Z
x dx
1 −1/2
√
=
u
du
2
25 + x2
= u1/2 + C
√
=
25 + x2 + C
So which one is faster?
139
5 sec(θ) tan(θ) dθ
4.4 Trigonometric Substitutions
Z
Example 4.7. Find
Brian E. Veitch
√
x3 9 − x2
Since the denominator has the form a2 − x2 , we use the substitution x = 3 sin(θ), and
dx = 3 cos(θ) dθ
Z
Z
q
√
3
2
x 9 − x dx =
27 sin (θ) 9 − 9 sin2 (θ) · (3 cos(θ)) dθ
Z
p
=
27 sin3 (θ) 9 cos2 (θ) · 3 cos(θ) dθ
Z
5
= 3
sin3 (θ) cos2 (θ) dθ
Z
5
= 3
sin2 (θ) cos2 (θ) · sin(θ) dθ
Z
5
= 3
(1 − cos2 (θ)) cos2 (θ) · sin(θ) dθ
3
Let u = cos(θ), and du = − sin(θ) dθ
5
3
Z
2
2
5
Z
(1 − cos (θ)) cos (θ) · sin(θ) dθ = −3
5
(1 − u2 )u2 du
Z
u2 − u4 du
1 5
5 1 3
u − u
= −3
3
5
3
cos5 (θ)
5 cos (θ)
= −3
−
3
5
= −3
Now we just have to figure out what cos(θ) is. To do that, we use the right triangle again.
Recall that x = 3 sin(θ)
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4.4 Trigonometric Substitutions
Brian E. Veitch
√
9 − x2
(9 − x2 )1/2
=
. Now we just replace the cos(θ) with this
We see that cos(θ) =
3
3
in our integral and we’re done!
√
3
(9 − x2 )3/2
9 − x2
3
=
.
Note cos (θ) =
3
33
Z
x
3
√
9−
x2
1
1
2 3/2
2 5/2
(9 − x ) −
(9 − x )
+C
dx = −3
34
5 · 35
(9 − x2 )3/2 = −
15 − (9 − x2 ) + C
5
(9 − x2 )3/2 2
= −
x +6 +C
5
5
The last few steps weren’t necessary, but it doesn’t hurt to practice simplifying.
Z
Example 4.8. Find
x2
√
1
x2 − 36
Let x = 6 sec(θ), and dx = 6 sec(θ) tan(θ)dθ.
Before setting up the integral, we can substitute and simplify
√
x2 − 36 =
√
p
36(sec2 (θ) − 9) = 6 tan(θ)
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x2 − 36 right now.
4.4 Trigonometric Substitutions
Brian E. Veitch
and
x2 = 36 sec2 (θ)
Now we substitute,
Z
Z
1
1
√
=
· 6 sec(θ) tan(θ) dθ
2
2
2
36 sec (θ) · 6 tan(θ)
x x − 36
Z
1
dθ
=
36 sec(θ)
Z
1
=
cos(θ) dθ
36
1
=
sin(θ) + C
36
We need to use the right triangle to rewrite sin(θ) in terms of x.
If x = 6 sec(θ), then
x
6
= sec(θ) → cos(θ) =
6
x
√
x2 − 36
x
√
Z
1
1
x2 − 36
√
Therefore,
=
·
+C
36
x
x2 x2 − 36
From here you can see sin(θ) =
142
4.4 Trigonometric Substitutions
Z
Example 4.9. Find
√
Brian E. Veitch
x2
dx
9 − x2
Since the denominator has the form a2 − x2 , we use the following substitution,
x = 3 sin(θ) , dx = 3 cos(θ)
To help simplify before we start the intergral,
√
Z
9−
x2
q
= 9 − 9 sin2 (θ) = 3 cos(θ)
Z
x
9 sin2 (θ)
√
· 3 cos(θ) dθ
dx =
3 cos(θ)
9 − x2
Z
=
9 sin2 (θ) dθ
Z
1
= 9
(1 − cos(2θ)) dθ
2
Z
9
1 − cos(θ) dθ
=
2
9
1
=
θ − sin(2θ) + C
2
2
9
9
=
θ − sin(2θ) + C
2
4
Now we use the right triangle,
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4.4 Trigonometric Substitutions
Brian E. Veitch
Note that we can’t use sin(2θ), since the angle is just a single θ. To deal with this, we
use the trig identity
sin(2θ) = 2 sin(θ) cos(θ)
Z
9
9
x2
√
dx = θ − sin(θ) cos(θ)
2
2
9 − x2
−1
Using the triangle, we have θ = sin
x
x
, sin(θ) = , and cos(θ) =
3
3
Final Answer:
x√ 9 − x2 9
x2
−1 x
√
dx =
sin
−
+C
2
3
9
9 − x2
Z
Z
Example 4.10. Find
x
√
dx
2
x + 4x + 5
144
√
9 − x2
3
4.4 Trigonometric Substitutions
Brian E. Veitch
The problem here is what’s in the square root doesn’t follow one of our three forms,
which are x2 − a2 , x2 + a2 , or a2 − x2 . So we need to complete the square on x2 + 4x + 5 so
it has one of these forms.
x2 + 4x + 5 = x2 + 4x+? −? + 5
The missing number ?, is ? =
−b
2a
2
=
−4
2(1)
2
= (−2)2 = 4.
x2 + 4x + 5 = x2 + 4x + 4 − 4 + 5
x2 + 4x + 5 = (x + 2)2 + 1
Now we have one of the correct forms. It has the form x2 + a2 , well sorta. But here’s
what we have now.
Z
x
√
dx =
2
x + 4x + 5
Z
We need to substitute.
1. Let u = x + 2 and x = u − 2
145
x
p
dx
(x + 2)2 + 1
4.4 Trigonometric Substitutions
Brian E. Veitch
2. du = dx
3. Substitute
Z
Z
x
p
dx =
(x + 2)2 + 1
u−2
√
du
u2 + 1
4. Now we do the trig substitution
Let x = 1 tan(θ), du = sec2 (θ) dθ.
√
Z
u2
q
p
+ 1 = tan2 (θ) + 1 = sec2 (θ) = sec(θ)
x
u−2
√
du
u2 + 1
Z
tan(θ) − 2
=
· sec2 (θ) dθ
sec(θ)
Z
=
sec(θ) tan(θ) − 2 sec(θ) dθ
Z
= sec(θ) − 2 sec(θ)
Z
p
dx =
(x + 2)2
= sec(θ) − 2 ln | sec(θ) + tan(θ)| + C
Now we use the right triangle.
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4.4 Trigonometric Substitutions
√
From here we get sec(θ) =
Brian E. Veitch
u2 + 1
and tan(θ) = u
1
sec(θ) − 2 ln | sec(θ) + tan(θ)| + C =
Z
√
√
√
u2 + 1 − 2 ln u2 + 1 + u + C
√
√
x
= x2 + 4x + 5 − 2 ln x2 + 4x + 5 + (x + 2) + C
x2 + 4x + 5
147