**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:

∫ *x*^{2} *dx* = ∫ *x · x dx* = ∫ *x dx* · ∫ *x dx*

because the left-hand side is *x* ^{3} /3 while the right-hand side is ( *x*^{2} /2) · ( *x*^{2} /2) = *x*^{4} /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,

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

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 = x*^{2} /2 and *du* = − sin *x dx* and the new integral

is more complicated.

**Example 2**

Calculate the integral

∫ *x*^{2} · *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*) = *x*^{2} *du = u′* ( *x* ) *dx* = 2 *x dx*

*v* (*x*) = *e ^{x} dv = v′* (

*x*)

*dx*=

*e*

^{x}dxThen the integration by parts formula tells us that

We see that we have transformed the integral into a simpler one (involving *x · e ^{x}* instead of

*x*

^{2}·

*e*), but another integration by parts will be required. Now we take

^{x}*u*= 2

*x*and

*dv = e*. Then

^{x}dx*u* ( *x* ) = 2*x du* = *u′* ( *x* ) *dx* = 2 *dx*

*v* ( *x* ) = *e ^{x} dv = v′* (

*x*)

*dx*=

*e*

^{x}dxSo equation (*) equals

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

**You Try It** : Calculate the integral

∫ *x*^{2} log *x dx* .

**Example 3**

Calculate

**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

**You Try It** : Evaluate

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

**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

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* = sin^{2} *x* .

Throwing in the usual constant of integration, we obtain

Now we complete our work by evaluating the definite integral:

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

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

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