6.2Integration bySubstitution & Separable Differential Equations
M.L.King Jr. Birthplace, Atlanta, GA Greg KellyHanford High SchoolRichland, WashingtonPhoto by Vickie Kelly, 2002
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.
Example 1:
52x dx Let 2u x
du dx5u du61
6u C
62
6
xC
The variable of integration must match the variable in the expression.
Don’t forget to substitute the value for u back into the problem!
Example:(Exploration 1 in the book)
21 2 x x dx One of the clues that we look for is if we can find a function and its derivative in the integrand.
The derivative of is .21 x 2 x dx
1
2 u du3
22
3u C
3
2 22
13
x C
2Let 1u x
2 du x dx
Note that this only worked because of the 2x in the original.Many integrals can not be done by substitution.
Example 2:
4 1 x dx Let 4 1u x
4 du dx
1
4du dx
Solve for dx.1
21
4
u du3
22 1
3 4u C
3
21
6u C
3
21
4 16
x C
Example: (Not in book)
2 3sin x x dx 3Let u x23 du x dx
21
3du x dx
We solve for because we can find it in the integrand.
2 x dx
1sin
3u du
1cos
3u C
31cos
3x C
Example 8:
24
0tan sec x x dx
The technique is a little different for definite integrals.
Let tanu x2sec du x dx
0 tan 0 0u
tan 14 4
u
1
0 u du
We can find new limits, and then we don’t have to substitute back.
new limit
new limit
12
0
1
2u
1
2We could have substituted back and used the original limits.
Example 8:
24
0tan sec x x dx
Let tanu x
2sec du x dx4
0 u du
Wrong!The limits don’t match!
42
0
1tan
2x
2
21 1tan tan 0
2 4 2
2 21 11 0
2 2
u du21
2u
1
2
Using the original limits:
Leave the limits out until you substitute back.
This is usually more work than finding new limits
Example: (Exploration 2 in the book)
1 2 3
13 x 1 x dx
3Let 1u x
23 du x dx 1 0u
1 2u 1
22
0 u du
23
2
0
2
3u Don’t forget to use the new limits.
3
22
23
22 2
3 4 2
3
Separable Differential Equations
A separable differential equation can be expressed as the product of a function of x and a function of y.
dyg x h y
dx
Example:
22dy
xydx
Multiply both sides by dx and divide
both sides by y2 to separate the
variables. (Assume y2 is never zero.)
22
dyx dx
y
2 2 y dy x dx
0h y
Separable Differential Equations
A separable differential equation can be expressed as the product of a function of x and a function of y.
dyg x h y
dx
Example:
22dy
xydx
22
dyx dx
y
2 2 y dy x dx
2 2 y dy x dx 1 2
1 2y C x C
21x C
y
2
1y
x C
2
1y
x C
0h y
Combined constants of integration
Example 9:
222 1 xdyx y e
dx
2
2
12
1xdy x e dx
y
Separable differential equation
2
2
12
1xdy x e dx
y
2u x
2 du x dx
2
1
1udy e du
y
1
1 2tan uy C e C 21
1 2tan xy C e C 21tan xy e C Combined constants of integration
Example 9:
222 1 xdyx y e
dx
21tan xy e C We now have y as an implicit
function of x.
We can find y as an explicit function
of x by taking the tangent of both sides.
21tan tan tan xy e C
2
tan xy e C
Notice that we can not factor out the constant C, because the distributive property does not work with tangent.