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The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions. We can sometimes use substitution to rewrite functions in a form that we can integrate.
1.
6.2
Integration by
Substitution &
Separable
Differential
Equations
Greg Kelly
M.L.King Jr. Birthplace, Atlanta, GA Hanford High School
Photo by Vickie Kelly, 2002 Richland, Washington
2.
The chain rule allows us to differentiate a wide variety
of functions, but we are able to find antiderivatives for
only a limited range of functions. We can sometimes
use substitution to rewrite functions in a form that we
can integrate.
3.
Example 1:
5
x 2 dx Let u x 2
du dx
5
u du The variable of integration
must match the variable in
1 6 the expression.
u C
6
6 Don’t forget to substitute the value
x 2 C
for u back into the problem!
6
4.
(Exploration 1 in the book) One of the clues that we look for is
2 if we can find a function and its
1 x 2 x dx derivative in the integrand.
2
The derivative of 1 x is 2 x dx .
1
2
u 2
du Let u 1 x
3 du 2 x dx
2
u C
2
3
Note that this only worked because
3 of the 2x in the original.
2
3
1 x 2 2
C Many integrals can not be done by
substitution.
5.
Example 2:
4 x 1 dx Let u 4 x 1
du 4 dx
1
1
Solve for dx.
u 2
du
4
1
du dx
3
4
2 1
u C
2
3 4
3
1
u C 2
6
1 3
4 x 1 2 C
6
6.
Example 3:
cos 7 x 5 dx Let u 7 x 5
du 7 dx
1
cos u 7 du 1
du dx
7
1
sin u C
7
1
sin 7 x 5 C
7
7.
Example: (Not in book)
x 2
sin x 3
dx Let u x 3
du 3 x 2 dx
1
sin u du 1
3 2
du x dx
3
1 2
We solve for x dx
cos u C
3 because we can find it
in the integrand.
1 3
cos x C
3
8.
Example 7:
4
x cos x dx
sin
4
sin x cos x dx Let u sin x
4
du cos x dx
u du
1 5
u C
5
1 5
sin x C
5
9.
Example 8:
The technique is a little different
for definite integrals.
2
tan
4
x sec x dx
0
new limit Let u tan x
1
u du
0
du sec 2 x dx
new limit We can find
1
u 0 tan 0 0 new limits,
1 2 and then we
u don’t have
2 0 u tan 1 to substitute
4 4
back.
1
2
We could have substituted back and
used the original limits.
10.
Example 8:
Using the original limits:
2
tan
4
x sec x dx
0
Let u tan x
u du
0
4
du sec 2 x dx
Leave the
u du limits out until
you substitute Wrong!
1 1
2
tan tan 0
2
back.
The
2 limits
4 don’t
2 match!
1 2
u
2 This is
1 usually
1 1 2 1 2
tan x
2 4 1 0 more work
2 2 2 than finding
2 0 new limits
11.
Example: (Exploration 2 in the book)
1
3x
2 3
x 1 dx Let u x 3 1 u 1 0
1
2
du 3x dx u 1 2
1
2
u
0
2
du
3 2
2 Don’t forget to use the new limits.
u 2
3 0
2 3
2 4 2
2 2 2 2
3 3 3
12.
Separable Differential Equations
A separable differential equation can be expressed as
the product of a function of x and a function of y.
dy
g x h y h y 0
dx
Example:
dy Multiply both sides by dx and divide
2 xy 2
dx both sides by y2 to separate the
variables. (Assume y2 is never zero.)
dy
2
2 x dx
y
y 2 dy 2 x dx
13.
Separable Differential Equations
A separable differential equation can be expressed as
the product of a function of x and a function of y.
dy
g x h y h y 0
dx
Example:
2
dy
dy 2 x dx
y
2 xy 2 1 2 Combined
dx y C1 x C2 constants of
dy 1 integration
2
2 x dx x 2 C
y y
1 1
y 2 dy 2 x dx 2 y y 2
x C x C
14.
Example 9:
dy x2
2 x 1 y e
2
Separable differential equation
dx
1 x2
2
dy 2 x e dx
1 y
1 x2 u x 2
1 y 2 dy 2 x e dx du 2 x dx
1 u
1 y 2 dy du
e
tan 1 y C1 eu C2
1 x2
tan y C1 e C2
1 x2
tan y e C Combined constants of integration
15.
Example 9:
dy x2
2 x 1 y e
2
dx
1 x2
tan y e C We now have y as an implicit
function of x.
x2
tan tan y tan e C We can find y as an explicit function
1
of x by taking the tangent of both
sides.
y tan e C x2
Notice that we can not factor out the constant C, because
the distributive property does not work with tangent.
16.
In another generation or so, we might be able to use
the calculator to find all integrals.
Until then, remember that half the AP exam and half the
nation’s college professors do not allow calculators.
You must practice finding integrals by hand until you are
good at it!
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