MATH 155 SUPPLEMENTAL NOTES 13

INTEGRATION BY PARTS

Suppose that we have an integral of the form

in which f (x) can be differentiated repeatedly and g(x) can be integrated repeatedly without difficulty. Here are a couple of examples of the type of integrals that I am talking about.

How do we integrate this type of integral? We will use a method called integration by parts, which comes from the product rule. Remember that we are talking about integrating an integral of a product, so let us derive the formula to do this.

Let u (x) and v (x) be functions. Let us look at the derivative of the product of these two functions.

Now, rewrite this formula.

Now, integrate both sides.

Here is the formula for the integration by parts.

You must chose on the two functions to be u and the other will be dv. How do you choose? First of all, the value of dv should be a function that you can easily integrate, and u should be easily differentiable. Second of all, if your choices for u and dv lead to a more complicated integral, then you have chosen incorrectly. So choose wisely!

EXAMPLE 1:

SOLUTION:

HINT: Any integral of the form of a polynomial times e ^x, let u be the polynomial and dv be e ^x.
u = x ² + 3x + 1	dv = e ^xdx
du = 2x + 3	v = e ^x

Is the new integral easier looking than the first? Yes! Can I integrate the new integral directly? No, so I will have to use integration by parts again. Hint: keep dv the same value as it was before!
u = 2x + 3		dv = e ^xdx
du = 2 dx		v = e ^x

Now, looking at the new integral, we should notice that we can integrate it directly.

EXAMPLE 2:

SOLUTION A:
Which of these two functions can be easily integrated? 4x is, so let it be dv.
u = sec ² 2x		dv = 4x dx
du = (2sec 2x)(sec 2x tan 2x)(2)dx		v = 2x ²

Well, look at this new integral. It is a mess, therefore we did not choose correctly.
SOLUTION B:
u = 4x	dv = sec ²2x dx
du = 4 dx

Now, the new integral is a straightforward integration. Let u = 2x, then du = 2 dx. Recall that the integral of the tan u is - ln \|cos u\|.

I purposely picked incorrectly in solution A to show you how the integral can become even harder than what you were given.

EXAMPLE 3:

SOLUTION:

Do we know how to integrate tan ^-1x? No, so let us let it be u. What will be dv? dv = dx.
u = tan ^-1x	dv = dx
	v = x

Now the new integral is a straightforward u-substitution. Let u = 1 + x ², then du = 2x dx.

EXAMPLE 4:

SOLUTION:

First of all, let us use a trig identity to replace the tan ² x. The reasons why we would like to do this is 1) we do not know how to integrate tan ² x directly, and 2) the derivative of tan ²x will lead us to a messier integral. Recall than tan ²x + 1 = sec ²x.

The first integral has to be done by using integration by parts, and the second integral is straightforward integration.
u = x	dv = sec ²x dx
du = dx	v = tan x

We will need the reference triangle for p / 3.

EXAMPLE 5:

SOLUTION:

Do we know how to integrate the ln x? No, so set u equal to the ln x.
u = ln x	dv = dx
	v = x

Sometimes we are given an integral in which we will have to solve for an unknown integral. I call these integrals circular integrals because we should end up back at the original integral.

EXAMPLE 6:

SOLUTION:

u = cos x	dv = e ^x dx
du = - sin x dx	v = e ^x

We will do integration by parts again. When we do this, we will keep dv = e ^xdx. If you do not, you will end up with zero on the right-hand side.
u = sin x	dv = e ^x dx
du = cos x dx	v = e ^x

Notice that we have the same integral on both sides, so we will move the integral on the right-hand side to the left-hand side.

Now solve for the original integral.

EXAMPLE 7:

SOLUTION:

u = sin 2x	dv = e ^-2xdx
du = 2cos 2x dx

u = cos 2x	dv = e ^-2xdx
du = -2sin 2x dx

EXAMPLE 8:

SOLUTION:

u = ln (x + x ²)	dv = dx
	v = x

Now we need to use long division to simplify the new integral.

Notice how I had to use one of the "tricks" from supplemental notes 12 in this problem and in an earlier problem in these notes. Those tricks are use full with any type of integration, and you will find that you will be using them quite frequently. If you are having problems with these "tricks" please refer back to supplemental notes 12. Work through these examples. The more of these types of integrals you do, the easier it will be to determine the values for u and for dv. If you have any questions or problems with any of these problems, feel free to contact me.

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