Integration by Inverse Substitution

diagram, we see ?
a2 ? x2
cos u =
a
And finally,
? ? ? ? ?
x?
a2 ? t2 dt = a2 u +
1
sin u cos u ? 0 = a2 sin?1(x/a)
+
1 ?x? ?
a2 ? x2
0 4 2 2 2 a a
x? a2 x 1 ?
a2 ? t2 dt =
2
sin?1(
a
) +
2x a2 ? x2
0
When the answer is this complicated, the route to getting there has to be rather complicated.
There’s no way to avoid the complexity.
1
Let’s double-check this answer. The area of the upper shaded sector in Figure 3 is a2u. The
2
area of the lower shaded region, which is a triangle of height
?
a2 ? x2 and base x, is 1x
?
a2 ? x2. 2
2
?
?
? =
? ? ?
Lecture 28 18.01 Fall 2006
0 x
u
Figure 3: Area divided into a sector and a triangle.
Here is a list of integrals that can be computed using a trig substitution and a trig identity.
integral substitution trig identity
dx
?
?
x2 + 1
x = tan u tan2 u + 1 = sec2 u
dx
?
x2
x = sec u sec2 u ? 1 = tan2 u
? ? 1
dx
?
1 ? x2
x = sin u 1 ? sin2 u = cos2 u
Let’s extend this further. How can we evaluate an integral like this?
dx
?
x2 + 4x
When you have a linear and a quadratic term under the square root, complete the square.
x2 + 4x = (something)2 ± constant
In this case,
(x + 2)2 = x2 + 4x + 4 =? x2 + 4x = (x + 2)2 ? 4
Now, we make a substitution.
v = x + 2 and dv = dx
Plugging these in gives us ? ?
dx dv
(x + 2)2 ? 4
?
v2 ? 4
Now, let
v = 2 sec u and dv = 2 sec u tan u
dv 2 sec u tan u du
?
v2
=
2 tan u
= sec u du
? 4
3
? ? ?
?
?
? ? ?
Lecture 28 18.01 Fall 2006
Remember that ?
sec u du = ln(sec u + tan u) + c
Finally, rewrite everything in terms of x.
2
v = 2 sec u ? cos u =
v
Set up a right triangle as in Figure 4. Express tan u in terms of v.
v
2
v²-4
u
Figure 4: sec u = v/2 or cos u = 2/v.
Just from looking at the triangle, we can read off
v
?
v2 ? 4
sec u = and tan u =
2 2
2 sec u du = ln v +
?
v2 ? 4
+ c 2 2
= ln(v + v2 ? 4) ? ln 2 + c
We can combine those last two terms into another constant, c?.
dx ?
?
x2 + 4x
= ln(x + 2 + x2 + 4x) +