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 3:

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: (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 7:

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 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.

Until then, remember that half the AP exam and half the nation’s college professors do not allow calculators.

In another generation or so, we might be able to use the calculator to find all integrals.

You must practice finding integrals by hand until you are good at it!