6.2 integration by substitution & separable differential equations
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6.2Integration bySubstitution & Separable Differential Equations
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:
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:
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:
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:
cos 7 5 x dx7 du dx
1
7du dx
1cos
7u du
1sin
7u C
1sin 7 5
7x C
Let 7 5u x
Example:
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:
4sin cos x x dx
Let sinu x
cos du x dx
4sin cos x x dx
4 u du51
5u C
51sin
5x C
Example:
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:
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:
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