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## Section5.6Integration by Parts

###### Motivating Questions
• How do we evaluate indefinite integrals that involve products of basic functions such as $\int x \sin(x) \, dx$ and $\int x e^x \, dx\text{?}$

• In the last section, we learned how to reverse the chain rule. How do we reverse other rules of differentiation? In this section, we'll tackle this question for the product rule of differentiation.

• What is the method of integration by parts and how can we consistently apply it to integrate products of basic functions?

• How does the algebraic structure of functions guide us in identifying $u$ and $dv$ in using integration by parts?

In Section5.5, we learned the technique of $u$-substitution for evaluating indefinite integrals. For example, the indefinite integral $\int x^3 \sin(x^4) \, dx$ is perfectly suited to $u$-substitution, because one factor is a composite function and the other factor is the derivative (up to a constant) of the inner function. Recognizing the algebraic structure of a function can help us to find its antiderivative.

Next we consider integrands with a different elementary algebraic structure: a product of basic functions. For instance, suppose we are interested in evaluating the indefinite integral

\begin{equation*} \int x \sin(x) \, dx\text{.} \end{equation*}

The integrand is the product of the basic functions $f(x) = x$ and $g(x) = \sin(x)\text{.}$ We know that it is relatively complicated to compute the derivative of the product of two functions, so we should expect that antidifferentiating a product should be similarly involved. Intuitively, we expect that evaluating $\int x \sin(x) \, dx$ will involve somehow reversing the Product Rule.

To that end, in Example5.45 we refresh our understanding of the Product Rule and then investigate some indefinite integrals that involve products of basic functions.

###### The Product Rule of Differentiation

In Section2.3, we developed the Product Rule and studied how it is employed to differentiate a product of two functions. In particular, recall that if $f$ and $g$ are differentiable functions of $x\text{,}$ then

\begin{gather*} \frac{d}{dx} \left[ f(x) \cdot g(x) \right] = f(x) \cdot g'(x) + g(x) \cdot f'(x) \text{.} \end{gather*}
###### Example5.45
1. For each of the following functions, use the Product Rule to find the function's derivative. Be sure to label each derivative by name (e.g., the derivative of $g(x)$ should be labeled $g'(x)$).

1. $g(x) = x\sin(x)$

2. $h(x) = xe^x$

3. $p(x) = x\ln(x)$

4. $q(x) = x^2 \cos(x)$

5. $r(x) = e^x \sin(x)$

1. $g'(x)=x\cos(x) + \sin(x)$

2. $h'(x)=xe^x + e^x$

3. $p'(x)=1+ \ln(x)$

4. $q'(x)=-x^2 \sin(x) + 2x\cos(x)$

5. $r'(x)=e^x \cos(x) + e^x \sin(x)$

Solution
1. \begin{align*} g'(x)&=x\cdot \frac{d}{dx}[\sin(x)] + \sin(x)\cdot \frac{d}{dx} [x] \\ &=x \cos(x)+ \sin(x) \end{align*}
2. \begin{align*} h'(x)&=x \cdot \frac{d}{dx}[e^x] + e^x \cdot \frac{d}{dx} [x] \\ &=xe^x + e^x \end{align*}
3. \begin{align*} p'(x)&=x \cdot \frac{d}{dx}[\ln(x)] + \ln(x) \cdot \frac{d}{dx} [x] \\ &=\frac{x}{x} + \ln (x) \\ &=1+ \ln(x) \end{align*}
4. \begin{align*} q'(x)&=x^2\cdot \frac{d}{dx} [\cos(x)] + \cos(x) \cdot \frac{d}{dx}[x^2] \\ &=-x^2 \sin(x) + 2x\cos(x) \end{align*}
5. \begin{align*} r'(x)&=e^x \cdot \frac{d}{dx}[\sin(x)] + \sin(x) \cdot \frac{d}{dx} [e^x] \\ &=e^x \cos(x) + e^x \sin(x) \end{align*}

1. $\int xe^x + e^x \, dx$

2. $\int e^x(\sin(x) + \cos(x)) \, dx$

3. $\int 2x\cos(x) - x^2 \sin(x) \, dx$

4. $\int x\cos(x) + \sin(x) \, dx$

5. $\int 1 + \ln(x) \, dx$

Hint

How do the integrands in part (b) compare to the answers in part (a)? What relationship holds between $h(x)$ from part a) ii. and the integrand from part b) i.?

1. $\int xe^x + e^x \, dx = xe^x + C$

2. $\int e^x(\sin(x) + \cos(x)) \, dx=e^x \sin(x) + C$

3. $\int 2x\cos(x) - x^2 \sin(x) \, dx =x^2 \cos(x) + C$

4. $\int x\cos(x) + \sin(x) \, dx=x\sin(x) + C$

5. $\int 1 + \ln(x) \, dx = x\ln(x) + C$

Solution
1. Since the derivative of $xe^x$ is $xe^x + e^x \text{,}$ then $xe^x +C$ is the general antiderivative of $xe^x + e^x \text{.}$ Thus, we know that $\int xe^x + e^x \, dx = xe^x + C\text{.}$ We can check our answer by differentiating $xe^x \text{,}$ which we did in part a), to make sure we get the integrand, $xe^x + e^x \text{.}$

2. Since the derivative of $e^x \sin(x)$ is $e^x(\sin(x) + \cos(x)) \text{,}$ then $e^x\sin(x) +C$ is the general antiderivative of $e^x(\sin(x) + \cos(x)) \text{.}$ Thus, we know that $\int e^x(\sin(x) + \cos(x)) \, dx=e^x \sin(x) + C\text{.}$

3. Since the derivative of $x^2\cos(x)$ is $2x\cos(x) -x^2\sin(x) \text{,}$ then $x^2\cos(x)+C$ is the general antiderivative of $2x\cos(x)-x^2\sin(x) \text{.}$ Thus, we know that $\int 2x\cos(x) - x^2 \sin(x) \, dx =x^2 \cos(x) + C\text{.}$

4. Since the derivative of $x\sin(x)$ is $x\cos(x) + \sin(x) \text{,}$ then $x\sin(x) +C$ is the general antiderivative of $x\cos(x)+ \sin(x) \text{.}$ Thus, we know that $\int x\cos(x) + \sin(x) \, dx=x\sin(x) + C\text{.}$

5. Since the derivative of $x\ln(x)$ is $1+\ln(x) \text{,}$ then $x\ln(x) +C$ is the general antiderivative of $1+ \ln(x) \text{.}$ Thus, we know that $\int 1 + \ln(x) \, dx = x\ln(x) + C \text{.}$

Observe that the examples in part (b) work nicely because of the derivatives you were asked to calculate in part (a). Each integrand in part (b) is precisely the result of differentiating one of the products of basic functions found in part (a). To see what happens when an integrand is still a product but not necessarily the result of differentiating a product of functions, we consider how to evaluate

\begin{equation*} \int x\cos(x) \, dx\text{.} \end{equation*}
###### Example5.46

Follow the steps outlined below to evlauate $\int x\cos(x) \, dx \text{.}$

1. First, observe that

\begin{equation*} \frac{d}{dx} \left[ x\sin(x) \right] = x\cos(x) + \sin(x)\text{.} \end{equation*}

Integrating both sides indefinitely and using the fact that the integral of a sum is the sum of the integrals, we find that

\begin{equation*} \int \left(\frac{d}{dx} \left[ x\sin(x) \right] \right) \, dx = \int x\cos(x) \, dx + \int \sin(x) \, dx\text{.} \end{equation*}

In this last equation, evaluate the indefinite integral on the left side of the equation as well as the rightmost indefinite integral on the right side of the equation.

Solution

From the Second Fundamental Theorem of Calculus5.4, we know that $\int \frac{d}{dx} \left[x\sin(x) \right] =x\sin(x) + C_1$ where $C_1$ is any constant. We also know that $\int \sin(x) \, dx = -\cos(x) + C_2$ where $C_2$ is any constant. Thus, we obtain the following equation:

\begin{gather*} x\sin(x) + C_1 = \int x \cos(x) \, dx - \cos(x) + C_2 \end{gather*}
2. In the most recent equation from (i.), solve the equation for the expression $\int x \cos(x) \, dx\text{.}$

Solution
\begin{align*} \int x \cos(x) \, dx&= x\sin(x) + \cos(x) + C_1-C_2 \\ \int x \cos(x) \, dx &= x\sin(x) + \cos(x) + C \end{align*}

where $C$ is any constant.

3. For which product of basic functions have you now found the antiderivative?

Solution

We found the antiderivative of the product of the functions $x$ and $\cos(x) \text{.}$

### SubsectionReversing the Product Rule: Integration by Parts

Example5.46 provides a clue to the general technique known as Integration by Parts, which comes from reversing the Product Rule. Recall that the Product Rule states that

\begin{equation*} \frac{d}{dx} \left[ f(x) g(x) \right] = f(x) g'(x) + g(x) f'(x)\text{.} \end{equation*}

Integrating both sides of this equation indefinitely with respect to $x\text{,}$ we find

\begin{equation} \int \frac{d}{dx} \left[ f(x) g(x) \right] \, dx = \int f(x) g'(x) \, dx + \int g(x) f'(x) \, dx\text{.}\label{E-intprod}\tag{5.6} \end{equation}

On the left side of Equation(5.6), we have the indefinite integral of the derivative of a function. Temporarily omitting the constant that may arise, we have

\begin{equation} f(x) g(x) = \int f(x) g'(x) \, dx + \int g(x) f'(x) \, dx\text{.}\label{E-intprod2}\tag{5.7} \end{equation}

We solve for the first indefinite integral on the left to generate the rule

\begin{equation} \int f(x) g'(x) \, dx = f(x) g(x) - \int g(x) f'(x) \, dx\text{.}\label{E-IBP1}\tag{5.8} \end{equation}

Often we express Equation(5.8) in terms of the variables $u$ and $v\text{,}$ where $u = f(x)$ and $v = g(x)\text{.}$ In differential notation, $du = f'(x) \, dx$ and $dv = g'(x) \, dx\text{,}$ so we can state the rule for Integration by Parts in its most common form as follows:

###### Integration by Parts Formula

If $u=f(x)$ and $v=g(x) \text{,}$ then $du=f'(x) \, dx$ and $dv=g'(x) \, dx \text{.}$ Then we can change Equation(5.8) to the following equation:

\begin{equation*} \int u \, dv = uv - \int v \, du\text{.} \end{equation*}

A way to remember the integration by parts formula is the following picture: Here, the curve $(u(x), v(x))$ cuts a rectangle in half. There are two ways to compute the area of the rectangle:

1. The first is the usual length times height. The length of the rectangle is $u\text{,}$ and the height is $v\text{,}$ so the area of the rectangle is $uv\text{.}$
2. The other way to find the area of the rectangle is to add the area under the curve and above the curve. The area under the curve comes from integrated $v(x)$ with respect to the horizontal axis (the $u$-axis), so the area under the curve is $\int v du\text{.}$ Similarly, the area above the curve is integrated with respect to the vertical axis, so its area is $\int u dv\text{.}$ Summing these gives the total area of the rectangle is $\int u dv + \int v du \text{.}$

Comparing the two methods tells us that $uv = \int u dv + \int v du \text{,}$ and rearranging this expression gives the integration by parts formula.

###### Method of Integration by Parts

To apply integration by parts to evaluate an indefinite integral, perform the following steps:

• Look for a product of basic functions that we can identify as $u$ and $dv\text{.}$ We need to antidifferentiate $dv$ to find $v\text{,}$ so keep this in mind when picking $u$ and $dv \text{.}$

• Once $u, v, du,$ and $dv$ have been identified, apply the integration by parts formula.
• If evaluating $\int v \, du$ is not more difficult than evaluating $\int u \, dv\text{,}$ then this substitution usually proves to be fruitful.

Note: there are some exceptions where $\int v\, du$ is just as difficult as $\int u \, dv$ but integration by parts is still helpful. We cover those cases in Using Integration by Parts Multiple Times.

To demonstrate this method, we consider the following example.

###### Example5.47

Evaluate the indefinite integral

\begin{equation*} \int x\cos(x) \, dx \end{equation*}

using integration by parts.

Solution

When we use integration by parts, we have a choice for $u$ and $dv\text{.}$ In this problem, we can either let $u = x$ and $dv = \cos(x) \, dx\text{,}$ or let $u = \cos(x)$ and $dv = x \, dx\text{.}$ While there is not a universal rule for how to choose $u$ and $dv\text{,}$ a good guideline is this: do so in a way that $\int v \, du$ is at least as simple as the original problem $\int u \, dv\text{.}$

This leads us to choose7Observe that if we considered the alternate choice, and let $u = \cos(x)$ and $dv = x \, dx\text{,}$ then $du = -\sin(x) \, dx$ and $v = \frac{1}{2}x^2\text{,}$ from which we would write $\int x\cos(x) \, dx = \frac{1}{2}x^2 \cos(x) - \int \frac{1}{2}x^2 (-\sin(x)) \, dx\text{.}$ Thus we have replaced the problem of integrating $x \cos(x)$ with that of integrating $\frac{1}{2}x^2 \sin(x)\text{;}$ the latter is clearly more complicated, which shows that this alternate choice is not as helpful as the first choice.$u = x$ and $dv = \cos(x) \, dx\text{,}$ from which it follows that $du = 1 \, dx$ and $v = \sin(x)\text{.}$ With this substitution, the rule for integration by parts tells us that

\begin{equation*} \int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \cdot 1 \, dx\text{.} \end{equation*}

All that remains to do is evaluate the (simpler) integral $\int \sin(x) \cdot 1 \, dx\text{.}$ Doing so, we find

\begin{equation*} \int x \cos(x) \, dx = x \sin(x) - (-\cos(x)) + C = x\sin(x) + \cos(x) + C\text{.} \end{equation*}

Observe that when we get to the final stage of evaluating the last remaining antiderivative, it is at this step that we include the integration constant, $+C\text{.}$

The general technique of integration by parts involves trading one integral for another since it converts the problem of evaluating $\int u \, dv$ to that of evaluating $\int v \, du\text{.}$ It is therefore important that our choice of $u$ and $v$ make the problem easier instead of harder! In Example5.47, the original integral to evaluate was $\int x \cos(x) \,dx\text{,}$ and through the substitution provided by integration by parts, we were instead able to evaluate $\int \sin(x) \cdot 1 \, dx\text{.}$ Note that the original function $x$ was replaced by its derivative, while $\cos(x)$ was replaced by its antiderivative.

###### Example5.48

Evaluate each of the following indefinite integrals. Check each antiderivative that you find by differentiating.

1. $\int te^{-t} \, dt$

2. $\int 4x \sin(3x) \, dx$

3. $\int z \sec^2(z) \,dz$

4. $\int x \ln(x) \, dx$

Hint
1. Try $u=t\text{.}$

2. Let $dv=\sin(3x)dx$

3. Remember that $\int \sec^2(z) \,dz = \tan(z)\text{.}$

4. Note that $\ln(x) \, dx$ has a simple derivative to work with.

1. $\int t e^{-t} dt = -te^{-t} - e^{-t} +C\text{.}$

2. $\int 4x \sin(3x) dx = -\dfrac{4}{3} x \cos(3x) + \frac{4}{9} \sin(3x) + C \text{.}$

3. $\int z \sec^2(z) dz = z \tan(z) + \ln |\cos(z)| + C \text{.}$

4. $\int x\ln(x) dx = \frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2 + C \text{.}$

Solution
1. Using $u = t$ and $dv = e^{-t} dt\text{,}$ we obtain $du = dt$ and $v = -e^{-t}\text{.}$ So

\begin{align*} \int t e^{-t} dt \amp = -te^{-t} + \int e^{-t} dt\\ \amp = -te^{-t} - e^{-t} + C \end{align*}

We check using differentiation as follows:

\begin{align*} \frac{d}{dt} \left( -te^{-t} - e^{-t} + C \right) \amp = \left( t e^{-t} - e^{-t} \right) + e^{-t}\\ \amp = te^{-t} \end{align*}
2. Using $u = 4x$ and $dv = \sin(3x) dx\text{,}$ we obtain $du = 4 dx$ and $v = -\dfrac{1}{3} \cos(3x)\text{.}$ So

\begin{align*} \int 4x \sin(3x) dx \amp = -\dfrac{4}{3} x \cos(3x) + \dfrac{4}{3} \int \cos(3x) dx\\ \amp = -\dfrac{4}{3} x \cos(3x) + \frac{4}{9} \sin(3x) + C \end{align*}

We check using differentiation as follows:

\begin{align*} \frac{d}{dx} \left( -\dfrac{4}{3} x \cos(3x) + \frac{4}{9} \sin(3x) + C \right) \amp = 4x \sin(3x) - \frac{4}{3} \cos(3x) + \frac{4}{3} \cos(3x)\\ \amp = 4x \sin(3x) \end{align*}
3. Using $u = z$ and $dv = \sec^2(z) dz\text{,}$ we obtain $du = dz$ and $v = \tan(z)\text{.}$ So

\begin{align*} \int z \sec^2(z) dz \amp = z \tan(z) - \int \tan(z) \,dz\\ \int z \sec^2(z) dz \amp = z \tan(z) - \int \frac{\sin(z)}{\cos(z)} \,dz\\ \amp = z \tan(z) + \ln |\cos(z)| + C \end{align*}

We check using differentiation as follows:

\begin{align*} \frac{d}{dz} \left( z \tan(z) + \ln |\cos(z)| + C \right) \amp = z\sec^2(z) + \tan(z) - \frac{\sin(z)}{\cos(z)}\\ \amp = z\sec^2(z) + \tan(z) - \tan(z)\\ \amp = z \sec^2(z) \end{align*}
4. Using $u = \ln(x)$ and $dv = x dx\text{,}$ we obtain $du = \dfrac{1}{x} dx$ and $v = \dfrac{1}{2}x^2\text{.}$ So

\begin{align*} \int x\ln(x) dx \amp = \frac{1}{2}x^2 \ln(x) - \frac{1}{2} \int x dx\\ \amp = \frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2 + C \end{align*}

We check using differentiation as follows:

\begin{align*} \frac{d}{dx} \left( \frac{1}{2}x^2 \ln(x) - \frac{1}{4}x^2 + C \right) \amp = \frac{1}{2}x + x \ln(x) - \frac{1}{2} x\\ \amp = x \ln(x) \end{align*}

### SubsectionSome Subtleties with Integration by Parts

Sometimes integration by parts is not an obvious choice, but the technique is appropriate nonetheless. Integration by parts allows us to replace one function in a product with its derivative while replacing the other with its antiderivative. For instance, consider evaluating

\begin{equation*} \int \arctan(x) \, dx\text{.} \end{equation*}

Initially, this problem seems ill-suited to integration by parts, since there does not appear to be a product of functions present. But if we note that $\arctan(x) = \arctan(x) \cdot 1\text{,}$ and realize that we know the derivative of $\arctan(x)$ as well as the antiderivative of $1\text{,}$ we see the possibility for the substitution $u = \arctan(x)$ and $dv = 1 \, dx\text{.}$ We explore this substitution further in Example5.49.

In a related problem, consider $\int t^3 \sin(t^2) \, dt\text{.}$ Observe that there is a composite function present in $\sin(t^2)\text{,}$ which normally suggests $u$-substitution, but there is not an obvious function-derivative pair, as we have $t^3$ (rather than simply $t$) multiplying $\sin(t^2)\text{.}$ In this problem we use both $u$-substitution and integration by parts. First we write $t^3 = t \cdot t^2$ and consider the indefinite integral

\begin{equation*} \int t \cdot t^2 \cdot \sin(t^2) \, dt\text{.} \end{equation*}

We let $z = t^2$ so that $dz = 2t \, dt\text{,}$ and thus $t \, dt = \frac{1}{2} \, dz\text{.}$ (We are using the variable $z$ to perform a $z$-substitution first so that we may then apply integration by parts.) Under this $z$-substitution, we now have

\begin{equation*} \int t \cdot t^2 \cdot \sin(t^2) \, dt = \int z \cdot \sin(z) \cdot \frac{1}{2} \, dz\text{.} \end{equation*}

The resulting integral can be evaluated by parts. This, too, is explored further in Example5.49.

These problems show that we sometimes must think creatively in choosing the variables for substitution in integration by parts, and that we may need to use substitution for an additional change of variables.

###### Example5.49

Evaluate each of the following indefinite integrals, using the provided hints.

1. Evaluate $\int \arctan(x) \, dx$ by using Integration by Parts with the substitution $u = \arctan(x)$ and $dv = 1 \, dx\text{.}$

2. Evaluate $\int \ln(z) \,dz\text{.}$ Consider a similar substitution to the one in (a).

3. Use the substitution $z = t^2$ to transform the integral $\int t^3 \sin(t^2) \, dt$ to a new integral in the variable $z\text{,}$ and evaluate that new integral by parts.

4. Evaluate $\int s^5 e^{s^3} \, ds$ using an approach similar to that described in (c).

5. Evaluate $\int e^{2t} \cos(e^t) \, dt\text{.}$ You will find it helpful to note that $e^{2t} = e^t \cdot e^t\text{.}$

1. $\int{\arctan(x) dx} = x\arctan(x) - \frac{1}{2} \ln \left( 1 + x^2 \right) + C \text{.}$

2. $\int \ln(z) dz = z \ln(z) - z + C \text{.}$

3. $\int t^3 \sin(t^2) dt = \frac{1}{2} \left( -t^2 \cos\left(t^2 \right) + \sin\left(t^2\right) \right) +C\text{.}$

4. $\int s^5 e^{s^3} ds = \frac{1}{3} \left( s^3 e^{s^3} - e^{s^3} \right) + C \text{.}$

5. $\int e^{2t} \cos\left( e^t \right) dt = e^t \sin \left( e^t \right) + \cos \left( e^t \right) + C \text{.}$

Solution
1. We use integration by parts with $u = \arctan(x)$ and $dv = dx\text{.}$ So $du = \dfrac{1}{1+x^2}dx$ and $v = x\text{.}$

\begin{align*} \int{\arctan(x) dx} \amp = uv - \int{v du}\\ \amp = x\arctan(x) - \int\frac{x}{1+x^2} dx \end{align*}

For the integral in the last equation, use the substitution $z = 1+x^2$ and $dz = 2x\, dx\text{.}$

\begin{align*} \int \arctan(x) dx \amp = x\arctan(x) -\int\frac{x}{1+x^2} dx\\ \amp = x\arctan(x) - \frac{1}{2} \int \frac{1}{z} dz\\ \amp = x\arctan(x) - \frac{1}{2} \ln(z) + C\\ \amp = x\arctan(x) - \frac{1}{2} \ln \left( 1 + x^2 \right) + C \end{align*}
2. Using $u = \ln(z)$ and $dv = dz\text{,}$ we obtain $du = \dfrac{1}{z} dz$ and $v = z\text{.}$ So

\begin{align*} \int \ln(z) dz \amp = z \ln(z) - \int dz\\ \amp = z \ln(z) - z + C \end{align*}
3. Using $z = t^2\text{,}$ we obtain $dz = 2t dt$ and $\dfrac{1}{2} dz = t dt\text{.}$ So

\begin{align*} \int t^3 \sin(t^2) dt \amp = \int t^2 \sin(t^2) \cdot t dt\\ \amp = \int z \sin(z) \frac{1}{2} dz\\ \amp = \frac{1}{2} \int z \sin(z) dz \end{align*}

Now using integration by parts with $u = z$ and $dv = \sin(z) dz\text{,}$ we obtain $du = dz$ and $v = -\cos(z)\text{.}$ So

\begin{align*} \int t^3 \sin(t^2) dt \amp = \frac{1}{2} \int z \sin(z) dz\\ \amp = \frac{1}{2} \left( -z \cos(z) + \int \cos(z) dz \right)\\ \amp = \frac{1}{2} \left( -z \cos(z) + \sin(z) \right) + C \end{align*}

Finally, we use $z = t^2$ and obtain

\begin{equation*} \int t^3 \sin(t^2) dt = \frac{1}{2} \left( -t^2 \cos\left(t^2 \right) + \sin\left(t^2\right) \right) +C\text{.} \end{equation*}
4. Use a substitution with $z = s^3$ and $dz = 3s^2 ds\text{.}$

\begin{align*} \int s^5 e^{s^3} ds \amp = \int s^2 s^3 e^{s^3} ds\\ \amp = \frac{1}{3} \int z e^z dz \end{align*}

Now use integration by parts with $u = z$ and $dv = e^z dz\text{.}$ So $du = dz$ and $v = e^z\text{.}$

\begin{align*} \int s^5 e^{s^3} ds \amp = \frac{1}{3} \int z e^z dz\\ \amp = \frac{1}{3} \left( z e^z - \int e^z dz \right)\\ \amp = \frac{1}{3} \left( z e^z - e^z \right) + C\\ \amp = \frac{1}{3} \left( s^3 e^{s^3} - e^{s^3} \right) + C \end{align*}
5. Using $z = e^t\text{,}$ we see that $dz = e^t dt$ and so

\begin{align*} \int e^{2t} \cos\left( e^t \right) dt \amp = \int e^t \cos\left( e^t \right) \cdot e^t dt\\ \amp = \int z \cos(z) dz \end{align*}

In Example5.47, we saw that

\begin{equation*} \int z \cos(z) dz = z \sin(z) + \cos(z) + C\text{.} \end{equation*}

Combining these two results, we obtain

\begin{align*} \int e^{2t} \cos\left( e^t \right) dt \amp = \int z \cos(z) dz\\ \amp = z \sin(z) + \cos(z) + C\\ \amp = e^t \sin \left( e^t \right) + \cos \left( e^t \right) + C \end{align*}

### SubsectionUsing Integration by Parts Multiple Times

Integration by parts is well suited to integrating the product of basic functions, allowing us to trade a given integrand for a new one where one function in the product is replaced by its derivative, and the other is replaced by its antiderivative. The goal in this trade of $\int u \, dv$ for $\int v \, du$ is that the new integral be simpler to evaluate than the original one. Sometimes it is necessary to apply integration by parts more than once in order to evaluate a given integral.

###### Example5.50

Evaluate $\int t^2 e^t \, dt\text{.}$

Solution

Let $u = t^2$ and $dv = e^t \, dt\text{.}$ Then $du = 2t \, dt$ and $v = e^t\text{,}$ and thus

\begin{equation*} \int t^2 e^t \, dt = t^2 e^t - \int 2t e^t \, dt\text{.} \end{equation*}

The integral on the right side is simpler to evaluate than the one on the left, but it still requires integration by parts. Now letting $u = 2t$ and $dv = e^t \, dt\text{,}$ we have $du = 2\, dt$ and $v = e^t\text{,}$ so that

\begin{equation*} \int t^2 e^t \, dt = t^2 e^t - \left( 2t e^t - \int 2 e^t \, dt \right)\text{.} \end{equation*}

(Note the parentheses, which remind us to distribute the minus sign to the entire value of the integral $\int 2t e^t \, dt\text{.}$) The final integral on the right is a basic one; evaluating that integral and distributing the minus sign, we find

\begin{equation*} \int t^2 e^t \, dt = t^2 e^t - 2t e^t + 2 e^t + C\text{.} \end{equation*}

Of course, even more than two applications of integration by parts may be necessary. In the preceding example, if the integrand had been $t^3e^t\text{,}$ we would have had to use integration by parts three times.

Next, we consider the slightly different scenario where multiple uses of integration by parts are necessary.

###### Example5.51

Evaluate $\int e^t \cos(t) \, dt\text{.}$

Hint

Apply integration by parts twice, then rearrange the equation to solve for $\int e^t \cos(t) \, dt\text{.}$

$\int e^t \cos(t) \, dt= \frac{1}{2}(e^t\cos(t)+e^t\sin(t)) +C\text{.}$

Solution

We can choose to let $u$ be either $e^t$ or $\cos(t)\text{;}$ we pick $u = \cos(t)\text{,}$ and thus $dv = e^t \, dt\text{.}$ With $du = -\sin(t) \, dt$ and $v = e^t\text{,}$ integration by parts tells us that

\begin{equation*} \int e^t \cos(t) \, dt = e^t \cos(t) - \int e^t (-\sin(t))\, dt\text{,} \end{equation*}

or equivalently that

\begin{equation} \int e^t \cos(t) \, dt = e^t \cos(t) + \int e^t \sin(t) \, dt\text{.}\label{E_IBPtwice}\tag{5.9} \end{equation}

The new integral has the same algebraic structure as the original one. While the overall situation isn't necessarily better than what we started with, it hasn't gotten worse. Thus, we proceed to integrate by parts again. This time we let $u = \sin(t)$ and $dv = e^t \, dt\text{,}$ so that $du = \cos(t) \, dt$ and $v = e^t\text{,}$ which implies

\begin{equation} \int e^t \cos(t) \, dt = e^t \cos(t) + \left( e^t \sin(t) - \int e^t \cos(t) \, dt \right)\text{.}\label{E_IBPtwice2}\tag{5.10} \end{equation}

We seem to be back where we started, as two applications of integration by parts has led us back to the original problem, $\int e^t \cos(t) \, dt\text{.}$ But if we look closely at Equation(5.10), we see that we can use algebra to solve for the value of the desired integral. Adding $\int e^t \cos(t) \, dt$ to both sides of the equation, we have

\begin{equation*} 2 \int e^t \cos(t) \, dt = e^t \cos(t) + e^t \sin(t)\text{,} \end{equation*}

and therefore

\begin{equation*} \int e^t \cos(t) \, dt = \frac{1}{2} \left( e^t \cos(t) + e^t \sin(t) \right) + C\text{.} \end{equation*}

Note that since we never actually encountered an integral we could evaluate directly, we didn't have the opportunity to add the integration constant $C$ until the final step.

###### Example5.52

Evaluate each of the following indefinite integrals.

1. $\int x^2 \sin(x) \, dx$

2. $\int t^3 \ln(t) \, dt$

3. $\int e^z \sin(z) \, dz$

4. $\int s^2 e^{3s} \, ds$

5. $\int t \arctan(t) \,dt$

Hint
1. Start with $u=x^2\text{.}$

2. Begin with $u=t^3\text{.}$

3. You'll have to integrate by parts twice.

4. Start with $u=s^2\text{.}$

5. Try $u=\arctan(x)\text{.}$ At a certain point in this problem, it is very helpful to note that $\frac{t^2}{1+t^2} = 1 - \frac{1}{1+t^2}\text{.}$ Just like while using partial fractions if the numerator and denominator of a rational function are the same, performing polynomial long division first can help.

1. \begin{gather*} \int x^2 \sin(x) dx = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C \text{.} \end{gather*}
2. \begin{gather*} \int t^3 \ln(t) dt = \frac{1}{4} t^4 \ln(t) - \frac{1}{16} t^4 + C \text{.} \end{gather*}
3. \begin{gather*} \int e^z \sin(z) dz = -\frac{1}{2}e^z \cos(z) + \frac{1}{2}e^z \sin(z) + C \text{.} \end{gather*}
4. \begin{gather*} \int s^2 e^{3s} ds = \frac{1}{3}s^2 e^{3s} - \frac{2}{9}s e^{3s} + \frac{2}{27} e^{3s} + C \text{.} \end{gather*}
5. \begin{align*} \int t \arctan(t) dt \amp= \frac{1}{2}t^2 \arctan(t) - \frac{1}{2}t - \frac{1}{2} \arctan(t) + C \text{.} \end{align*}
Solution
1. First, using $u = x^2$ and $dv = \sin(x) dx\text{,}$ we get $du = 2x dx$ and $v = -\cos(x)\text{.}$ So

\begin{equation*} \int x^2 \sin(x) dx = -x^2 \cos(x) + 2 \int x \cos(x) dx\text{.} \end{equation*}

Now, for $\int x \cos(x) dx\text{,}$ we use $u = x$ and $dv = \cos(x)$ and get $du = dx$ and $v =\sin(x)\text{.}$ So

\begin{align*} \int x^2 \sin(x) dx \amp = -x^2 \cos(x) + 2 \left( x \sin(x) - \int \sin(x) dx \right)\\ \amp = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C \end{align*}
2. Using $u = \ln(t)$ and $dv = t^3 dt\text{,}$ we get $du = \dfrac{1}{t} dt$ and $v = \dfrac{1}{4} t^4\text{.}$ So

\begin{align*} \int t^3 \ln(t) dt \amp = \frac{1}{4} t^4 \ln(t) - \frac{1}{4} \int t^3 dt\\ \amp = \frac{1}{4} t^4 \ln(t) - \frac{1}{4} \cdot \frac{1}{4} t^4 + C\\ \amp = \frac{1}{4} t^4 \ln(t) - \frac{1}{16} t^4 + C \end{align*}
3. We use $u = e^z$ and $dv = \sin(z) dz\text{.}$ This gives $du = e^z dz$ and $v = -\cos(z)\text{,}$ and

\begin{equation*} \int e^z \sin(z) dz = -e^z \cos(z) + \int e^z \cos(z) dz\text{.} \end{equation*}

For $\int e^z \cos(z) dz\text{,}$ we use $u = e^z$ and $dv = \cos(z) dz\text{.}$ So $du = e^z dz$ and $v = \sin(z)\text{,}$ and

\begin{align*} \int e^z \sin(z) dz \amp = -e^z \cos(z) + e^z \sin(z) dz - \int e^z \sin(z) dz\\ 2 \int e^z \sin(z) dz \amp = -e^z \cos(z) + e^z \sin(z) + C'\\ \int e^z \sin(z) dz \amp = -\frac{1}{2}e^z \cos(z) + \frac{1}{2}e^z \sin(z) + C \end{align*}
4. We use $u = s^2\text{,}$ $dv = e^{3s} ds\text{,}$ $du = 2s ds\text{,}$ and $v = \frac{1}{3} e^{3s}\text{.}$ This gives

\begin{equation*} \int s^2 e^{3s} ds = \frac{1}{3}s^2 e^{3s} - \frac{2}{3} \int s e^{3s} ds\text{.} \end{equation*}

We now use $u = s\text{,}$ $dv = e^{3s} ds\text{,}$ $du = ds\text{,}$ and $v = \dfrac{1}{3} e^{3s}$ for $\int s e^{3s} ds\text{.}$

\begin{align*} \int s^2 e^{3s} ds \amp = \frac{1}{3}s^2 e^{3s} - \frac{2}{3} \left( \frac{1}{3}s e^{3s} - \frac{1}{3}\int e^{3s} ds \right)\\ \amp = \frac{1}{3}s^2 e^{3s} - \frac{2}{3} \left( \frac{1}{3}s e^{3s} - \frac{1}{9} e^{3s} \right) + C\\ \amp = \frac{1}{3}s^2 e^{3s} - \frac{2}{9}s e^{3s} + \frac{2}{27} e^{3s} + C \end{align*}
5. Using $u = \arctan(t)\text{,}$ $dv = t dt\text{,}$ $du = \dfrac{1}{1+t^2} dt\text{,}$ and $v = \dfrac{1}{2} t^2\text{,}$ we get

\begin{align*} \int t \arctan(t) dt \amp = \frac{1}{2}t^2 \arctan(t) - \frac{1}{2} \int \frac{t^2}{1 + t^2} dt\\ \amp = \frac{1}{2}t^2 \arctan(t) - \frac{1}{2} \int \left( 1 - \frac{1}{1 + t^2} \right) dt\\ \amp = \frac{1}{2}t^2 \arctan(t) - \frac{1}{2}t - \frac{1}{2} \arctan(t) + C \end{align*}

### SubsectionEvaluating Definite Integrals Using Integration by Parts

We can use the technique of integration by parts to evaluate a definite integral by finding the indefinite integral and then plugging in the endpoints.

###### Example5.53

Evaluate the following definite integrals using the technique of integration by parts:

1. $\displaystyle \int_0^{\pi/2} t\sin(t) \, dt$
2. $\displaystyle \int_1^{2} te^{2t} \, dt$
Hint
1. Take $du=\sin(t)dt\text{.}$ What is $v\text{?}$
2. Take $v=e^{2t}\text{.}$ What is $u\text{?}$
1. $1$
2. $\sim 39.1$
Solution
1. One option is to find an antiderivative (using indefinite integral notation) and then apply the Fundamental Theorem of Calculus to find that

\begin{align*} \int_0^{\pi/2} t\sin(t) \, dt =\mathstrut \amp \left( -t \cos(t) + \sin(t) \right) \bigg\vert_0^{\pi/2}\\ =\mathstrut \amp \left( -\frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) \right) - \left( -0 \cos(0) + \sin(0) \right)\\ =\mathstrut \amp 1\text{.} \end{align*}

Alternatively, we can apply integration by parts and work with definite integrals throughout. With this method, we must remember to evaluate the product $uv$ over the given limits of integration. Using the substitution $u = t$ and $dv = \sin(t) \, dt\text{,}$ so that $du = dt$ and $v = -\cos(t)\text{,}$ we write

\begin{align*} \int_0^{\pi/2} t\sin(t) \, dt =\mathstrut \amp -t \cos(t) \bigg\vert_0^{\pi/2} - \int_0^{\pi/2} (-\cos(t)) \, dt\\ =\mathstrut \amp -t \cos(t) \bigg\vert_0^{\pi/2} + \sin(t) \bigg\vert_0^{\pi/2}\\ =\mathstrut \amp \left( -\frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) \right) - \left( -0 \cos(0) + \sin(0) \right)\\ =\mathstrut \amp 1\text{.} \end{align*}
2. We can take $u=\frac 1 2 t$ and $v=e^{2t}\text{,}$ so that $du =\frac 1 2 dt$ and $dv=2e^{2t}dt\text{.}$ Then

\begin{align*} \int_1^2 u\, dv =\mathstrut\amp \left. uv \right |_{t=1}^{t=2} - \int_1^2 v \, du \\ \int_1^2 te^{2t} \, dt = \mathstrut \amp \left. (\frac 12 t)(e^{2t}) \right |_{t=1}^{t=2} - \int_1^2 (e^{2t}) (\frac 1 2 dt) \\ = \mathstrut \amp \left. (\frac 12 t)(e^{2t}) \right |_{t=1}^{t=2}- \left. \frac 14 (e^{2t}) \right |_{t=1}^{t=2}\\ = \mathstrut \amp ((\frac 12 (2)))(e^{2(2))})-(\frac 12 (1))(e^{2(1)}))- (\frac 14 (e^{2(2)})-\frac 14 (e^{2(1)}))\\ = \mathstrut \amp \frac 34 e^4 - \frac 1 4 e^2 \approx 39.1\text{.} \end{align*}

As with any substitution technique, it is important to use notation carefully and completely, and to ensure that the end result makes sense.

### SubsectionWhen $u$-substitution and Integration by Parts Fail to Help

Both integration techniques we have discussed apply in relatively limited circumstances. It is not hard to find examples of functions for which neither technique produces an antiderivative; indeed, there are many, many functions that appear elementary but that do not have an elementary algebraic antiderivative. For instance, neither $u$-substitution nor integration by parts proves fruitful for the indefinite integrals

\begin{equation*} \int e^{x^2} \, dx \ \ \text{and} \ \ \int x \tan(x) \, dx\text{.} \end{equation*}

While there are other integration techniques, some of which we will consider briefly, none of them enables us to find an algebraic antiderivative for $e^{x^2}$ or $x \tan(x)\text{.}$ We do know from the Second Fundamental Theorem of Calculus that we can construct an integral antiderivative for each function; $F(x) = \int_0^x e^{t^2} \, dt$ is an antiderivative of $f(x) = e^{x^2}\text{,}$ and $G(x) = \int_0^{x} t \tan(t) \, dt$ is an antiderivative of $g(x) = x \tan(x)\text{.}$ But finding an elementary algebraic formula that doesn't involve integrals for either $F$ or $G$ turns out not only to be impossible through $u$-substitution or integration by parts, but indeed impossible altogether. Antidifferentiation is much harder in general than differentiation.

### SubsectionSummary

• Through the method of integration by parts, we can evaluate indefinite integrals, such as $\int x \sin(x) \, dx$ and $\int x \ln(x) \, dx\text{,}$ that involve products of basic functions. Using a substitution enables us to trade one of the functions in the product for its derivative, and the other for its antiderivative, in an effort to find a different product of functions that is easier to integrate.

• If the algebraic structure of an integrand is a product of basic functions in the form $\int f(x) g'(x) \, dx\text{,}$ we can use the substitution $u = f(x)$ and $dv = g'(x) \,dx$ and apply the rule

\begin{equation*} \int u \, dv = uv - \int v \, du \end{equation*}

to evaluate the original integral $\int f(x) g'(x) \, dx$ by instead evaluating

\begin{equation*} \int v \, du = \int f'(x) g(x) \, dx\text{.} \end{equation*}
• When deciding to integrate by parts, we have to select both $u$ and $dv\text{.}$ That selection is guided by the overall principle that the new integral $\int v \, du$ not be more difficult than the original integral $\int u \, dv\text{.}$ In addition, it is often helpful to recognize if one of the functions present is much easier to differentiate than antidifferentiate (such as $\ln(x)$), in which case that function often is best assigned the variable $u\text{.}$ In addition, $dv$ must be a function that we can antidifferentiate.

• Sometimes, integration by parts must be used more than once, or with another technique like $u$-substitution.

### SubsectionExercises

Let $f(t) = te^{-2t}$ and $F(x) = \int_0^x f(t) \, dt\text{.}$

1. Determine $F'(x)\text{.}$

2. Use the First FTC to find a formula for $F$ that does not involve an integral.

3. Is $F$ an increasing or decreasing function for $x \gt 0\text{?}$ Why?

Consider the indefinite integral given by $\int e^{2x} \cos(e^x) \, dx\text{.}$

1. Noting that $e^{2x} = e^x \cdot e^x\text{,}$ use the substitution $z = e^{x}$ to determine a new, equivalent integral in the variable $z\text{.}$

2. Evaluate the integral you found in (a) using an appropriate technique.

3. How is the problem of evaluating $\int e^{2x} \cos(e^{2x}) \, dx$ different from evaluating the integral in (a)? Do so.

4. Evaluate each of the following integrals as well, keeping in mind the approach(es) used earlier in this problem:

• $\int e^{2x} \sin(e^x) \, dx$

• $\int e^{3x} \sin(e^{3x}) \, dx$

• $\int xe^{x^2} \cos(e^{x^2}) \sin(e^{x^2}) \, dx$

For each of the following indefinite integrals,

1. Determine whether you would use $u$-substitution, integration by parts, neither*, or both to evaluate the integral.

2. In each case, write one sentence to explain your reasoning, and include a statement of any substitutions used. (That is, if you decide in a problem to let $u = e^{3x}\text{,}$ you should state that, as well as that $du = 3e^{3x} \, dx\text{.}$)

3. Use your chosen approach to evaluate each integral. (* one of the following problems does not have an elementary antiderivative and you are not expected to actually evaluate this integral; this will correspond with a choice of neither among those given.)

1. $\int x^2 \cos(x^3) \, dx$

2. $\int x^5 \cos(x^3) \, dx$ (Hint: $x^5 = x^2 \cdot x^3$)

3. $\int x\ln(x^2) \, dx$

4. $\int \sin(x^4) \, dx$

5. $\int x^3 \sin(x^4) \, dx$

6. $\int x^7 \sin(x^4) \, dx$

Use integration by parts and the trigonometric identity $\sin^2(\theta)+\cos^2(\theta)=1$ to find the following integrals:

1. $\int \sin^2(t)\, dt$
2. $\int \sin(t) \cos(t)\, dt$

(Note: these integrals can also be solved by other methods, but for this problem try integration by parts.)

Use integration by parts to find the following integrals:

1. $\int t e^t\, dt$
2. $\int t^2 e^t\, dt$
3. $\int t^3 e^t\, dt$
4. $\int t^4 e^t\, dt$
5. Do you notice a pattern in your answers? Without finding the integral, what do you think will be $\int t^5 e^t\, dt \text{?}$ Then check your answer online or by performing another integration by parts.

(Note: these integrals can also be solved by other methods, but for this problem try integration by parts.)