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Integration by Parts Help

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Introduction to Integration by Parts

We know that the integral of the sum of two functions is the sum of the respective integrals. But what of the integral of a product? The following reasoning is incorrect:

x2 dx = ∫ x · x dx = ∫ x dx · ∫ x dx

because the left-hand side is x 3 /3 while the right-hand side is ( x2 /2) · ( x2 /2) = x4 /4.

The correct technique for handling the integral of a product is a bit more subtle, and is called integration by parts . It is based on the product rule

( u · v )′ = u′ · v + u · v′ .

Integrating both sides of this equation, we have

∫ ( u · v )′ dx = ∫ u′ · v dx + ∫ u · v′ dx .

The Fundamental Theorem of Calculus tells us that the left-hand side is u · v . Thus

u · v = ∫ u′ · v dx + ∫ u · v′ dx

or

u · v′ dx = u · v − ∫ v · u′ dx .

It is traditional to abbreviate u′ ( x ) dx = du and v′ (x) dx = dv . Thus the integration by parts formula becomes

u d v = uv − ∫ v d u .

Let us now learn how to use this simple new formula.

Example 1

Calculate

x · cos x dx .

Solution 1

We observe that the integrand is a product. Let us use the integration by parts formula by setting u ( x ) = x and dv = cos x dx . Then

u (x) = x du = u′ (x) dx = 1 dx = dx

v (x) = sin x dv = v′ (x) dx = cos x dx

Of course we calculate v by anti-differentiation.

According to the integration by parts formula,

Methods of Integration 7.1 Integration by Parts

Math Note : Observe that we can check the answer in the last example just by differentiation:

Methods of Integration 7.1 Integration by Parts

The choice of u and v in the integration by parts technique is significant. We selected u to be x because then du will be 1 dx , thereby simplifying the integral. If we had instead selected u = cos x and dv = x dx then we would have found that v = x2 /2 and du = − sin x dx and the new integral

Methods of Integration 7.1 Integration by Parts

is more complicated.

Example 2

Calculate the integral

x2 · e x dx .

Solution 2

Keeping in mind that we want to choose u and v so as to simplify the integral, we take u = x 2 and dv = e x dx . Then

u (x) = x2 du = u′ ( x ) dx = 2 x dx

v (x) = e x dv = v′ ( x ) dx = ex dx

Then the integration by parts formula tells us that

Methods of Integration 7.1 Integration by Parts

We see that we have transformed the integral into a simpler one (involving x · e x instead of x 2 · e x ), but another integration by parts will be required. Now we take u = 2 x and dv = e x dx . Then

u ( x ) = 2x du = u′ ( x ) dx = 2 dx

v ( x ) = e x dv = v′ ( x ) dx = e x dx

So equation (*) equals

Methods of Integration 7.1 Integration by Parts

We leave it to the reader to check this last answer by differentiation.

You Try It : Calculate the integral

x2 log x dx .

Example 3

Calculate

Methods of Integration 7.1 Integration by Parts

Solution 3

This example differs from the previous ones because now we are evaluating a definite integral (i.e., an integral with numerical limits). We still use the integration by parts formula, keeping track of the numerical limits of integration.

We first notice that, on the one hand, the integrand is not a product. On the other hand, we certainly do not know an antiderivative for log x . We remedy the situation by writing log x = 1 · log x . Now the only reasonable choice is to take u = log x and dv = 1 dx . Therefore

u ( x ) = log x du = u′ ( x ) dx = (1/ x ) dx

v ( x ) = x dv = v′ ( x ) dx = 1 dx

and

Methods of Integration 7.1 Integration by Parts

You Try It : Evaluate

Methods of Integration 7.1 Integration by Parts

We conclude this section by doing another definite integral, but we use a slightly different approach from that in Example 3.

Example 4

Calculate the integral

Methods of Integration 7.1 Integration by Parts

Solution 4

We use integration by parts, but we apply the technique to the corresponding indefinite integral. We let u = sin x and dv = cos x dx . Then

u ( x ) = sin x du = u′ ( x ) dx = cos x dx

v ( x ) = sin x dv = v′ ( x ) dx = cos x dx

So

Methods of Integration 7.1 Integration by Parts

At first blush, it appears that we have accomplished nothing. For the new integral is just the same as the old integral. But in fact we can move the new integral (on the right) to the left-hand side to obtain

2 ∫ sin x cos x dx = sin2 x .

Throwing in the usual constant of integration, we obtain

Methods of Integration 7.1 Integration by Parts

Now we complete our work by evaluating the definite integral:

Methods of Integration 7.1 Integration by Parts

We see that there are two ways to treat a definite integral using integration by parts. One is to carry the limits of integration along with the parts calculation. The other is to do the parts calculation first (with an indefinite integral) and then plug in the limits of integration at the end. Either method will lead to the same solution.

You Try It : Calculate the integral

Methods of Integration 7.1 Integration by Parts

Find practice problems and solutions for these concepts at: Methods Of Integration Practice Test.

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