1.2 Integration by Parts

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We will compute integrals using the integration by parts technique.

1 INTRODUCTION

In this section, we are interested in integrating products. We have some experience with this from our exposure to $u$ -substitution problems such as $\int 2x \cos (1 + x^2) \; dx.$ In this scenario, the factor of $2x$ in the integrand is of great significance as it is the derivative of $1 + x^2$ which is the inside of the composition $\cos (1+x^2)$ . We will now turn to problems where the factors in the product making up the integrand are unrelated in this or any other significant way, i.e., they are functionally independent of one another. We will effectively create a product rule for integrals and interestingly, we will use the product rule for derivatives to help us. One stark difference between this product rule for integrals and the product rule for derivatives, known as integration by parts (IBP), is that IBP does not give us a final answer. Instead, integration by parts simply transforms our problem into another, hopefully easier one, which we then have to solve. In this last respect, IBP is similar to $u$ -substitution. The following integrals can be computed using IBP:

$\int 3x\sin (4x) \; dx,\;\; \int x^2 e^{3x} \; dx,\;\; \int \ln (x) \; dx \;\;\text {and}\;\; \int e^{2x}\cos (3x) \; dx.$

2 IBP Formula

Recall the product rule: $(uv)' = u'v+uv'$ (where $u$ and $v$ are functions of $x$ ). Integrating both sides of this equation with respect to $x$ gives $\int (uv)' = \int u'v + \int uv',$ where the $dx$ has intentionally been suppressed from the notation for simplicity. Now the integral on the left hand side is simply $uv + C$ by the anti-differentiation definition of the integral symbol. So, $\int u'v + \int uv' = uv + C.$ Subtracting the second integral from both sides gives $\int uv' = uv - \int u'v,$ which can be condensed as $\int u \; dv = uv - \int v \; du,$ the famous integration by parts formula.

Integration by Parts $\int u \; dv = uv - \int v \; du$

1.: To use the IBP formula, the original problem must be written in the form $\int u \; dv$ . This means that $u$ and $dv$ must be declared (similar to declaring $u$ in $u$ -substitution), and then we compute $du$ by differentiating $u$ , and $v$ by anti-differentiating $dv$ .
2.: The constant of integration does not appear on the right hand side of the IBP formula (it mysteriously disappeared in the above derivation) because it was taken into account by the constant of integration implied in the integral on the right hand side, $\int v \,du$ .
3.: The IBP formula does not give a final answer to a problem. It replaces one integral with another another integral (which still has to be computed).

We will now use the integration by parts (IBP) formula to compute integrals involving products of unrelated factors, like the ones shown above.

example 1 Compute the integral using IBP: $\int x\sin (x) \;dx.$

The integrand is the product of two factors, $x$ and $\sin (x)$ . We consider one factor to differentiate and the other factor to anti-differentiate. The factor we prefer to differentiate is $x$ since the derivative of $x$ is a constant and constants are easy to deal with in integrals. As for the $\sin (x)$ factor, whether we differentiate it or anti-differentiate it makes little difference. In fact, only a change in sign occurs, as $\frac {d}{dx}{\sin (x)} = \cos (x) \;\; \text { and }\;\; \int \sin (x) \;dx = -\cos (x) + C.$ Thus to apply the IBP formula, we let $u=x \;\;\text { and } \;\; dv = \sin (x) \;dx.$ Next, we compute $du$ by differentiating $u$ and we compute $v$ by anti-differentiating $dv$ : $du = dx \;\; \text { and }\;\; v = -\cos (x).$

We do not need to add the constant of integration to v, as any anti-derivative will suffice for our purpose and we choose the constant of integration to be zero.

Now we use the Integration by Parts formula, $\int u\;dv = uv-\int v\; du$ : \begin{align*} \int x\sin (x) &= \overbrace{x}^{u}\overbrace{[-\cos (x)]}^{v} - \int \overbrace{-\cos (x)}^{v} \;\overbrace{dx}^{du}\\ &= -x\cos (x) + \int \cos (x) \;dx\\ &= -x\cos (x) + \sin (x) + C. \end{align*}

We have found an answer, but at this stage we should prove that it is the correct answer by differentiating it. The result should be $x\sin (x)$ . Note that the differentiation requires the product rule on the $uv$ term: $-x\cos (x)$ . We have $\frac {d}{dx}{\left [-x\cos (x) + \sin (x) + C\right ]} = [-\cos (x) + x\sin (x)] + \cos (x) = x\sin (x)$ which proves that our answer was correct.

(problem 1a) Compute the integral using IBP: $\int xe^x \;dx.$

Let $u = \answer {x}\;\;$ and $\;\; dv = \answer {e^x dx}$ .
Then $du = \answer {dx}\;\;$ and $\;\; v = \answer {e^x}$ .
Thus $\int xe^x dx = xe^x - \int \answer {e^x} dx = \answer {xe^x - e^x} + C.$

(problem 1b) Compute the integral using IBP: $\int x\cos (x) \;dx.$

Let $u = \answer {x}\;\;$ and $\;\;dv = \answer {\cos (x) \;dx}$ .
Then $du = \answer {dx}\;\;$ and $\;\;v = \answer {\sin (x)}$ .
Thus $\int x\cos (x) dx = \answer {x\sin (x) + \cos (x)} + C.$

example 2 Compute the integral using IBP: $\int 3x\sin (4x) \;dx.$

Let $u = 3x\;\;$ and $\;\;dv = \sin (4x) \,dx$ .
Then $du = 3 \,dx\;\;$ and $\;\; v = -\frac 14 \cos (4x)$ . Next, the IBP formula gives \begin{align*} \int 3x\sin (4x) \;dx &= \overbrace{3x}^{u}\overbrace{\left [-\frac 14 \cos (4x)\right ]}^{v} - \int \overbrace{-\frac 14 \cos (4x)}^{v} \cdot \overbrace{3 \; dx}^{du} \\ &= -\frac 34 x\cos (4x) + \frac 34 \int \cos (4x) \; dx \\ &= -\frac 34 x\cos (4x) + \frac 34 \cdot \frac 14 \sin (4x) + C\\ &= -\frac 34 x\cos (4x) + \frac{3}{16}\sin (4x) + C. \end{align*}

(problem 2a) Compute the indefinite integral: $\int 2x\sin (5x)\;dx.$

Let $u = \answer {2x}\;\;$ and $\;\;dv = \answer {\sin (5x) dx}$ .
Then $du = \answer {2 dx}\;\;$ and $\;\; v = \answer {-\frac 15 \cos (5x)}$ .
Thus $\int 2x\sin (5x) dx = -\frac 25 x\cos (5x) - \int \answer {-\frac 25 \cos (5x)} dx = \answer {-\frac 25 x\cos (5x) +\frac {2}{25} \sin (5x)} + C.$

(problem 2b) Compute the indefinite integral: $\int 6xe^{-3x} \;dx.$ Let $u = \answer {6x}\;\;$ and $\;\;dv = \answer {e^{-3x} dx}$ .
Then $du = \answer {6 dx}\;\;$ and $\;\;v = \answer {-\frac 13 e^{-3x}}$ .
Thus $\int 6xe^{-3x} dx = -2 xe^{-3x} -\int \answer {-2 e^{-3x}} dx = \answer {-2 xe^{-3x} - \frac 23 e^{-3x}} + C.$

(problem 2c) Compute the integral using IBP: $\int x\sin (x/2) \;dx.$

Let $u = \answer {x}\;\;$ and $\;\;dv = \answer {\sin (x/2) dx}$ .
Then $du = \answer { dx}\;\;$ and $\;\;v = \answer {-2 \cos (x/2)}$ .
Thus $\int x\sin (x/2) dx = -2x\cos (x/2) - \int \answer {-2 \cos (x/2)} dx = \answer {-2x\cos (x/2) + 4 \sin (x/2)} + C.$

example 3, IBP twice Compute the integral: $\int x^2\sin (3x) \;dx.$

$\text {Let} \;\; u = x^2 \;\; \text { and } \;\; dv = \sin (3x) \,dx,$ $\text {then} \;\; du = 2x \,dx \;\; \text { and } \;\; v = -\frac 13 \cos (3x).$

Next, the IBP formula gives \begin{align*} \int x^2\sin (3x) \;dx &= x^2\left [-\frac 13 \cos (3x)\right ] - \int -\frac 13 \cos (3x) \cdot 2x \; dx \\ &= -\frac 13 x^2\cos (3x) + \frac 13 \int 2x\cos (3x) \; dx\\ \text{we now use IBP} & \text{ again with } \;\; u = 2x \;\text{ and } \; dv = \cos (3x) \, dx\\ & \qquad \qquad \quad du = 2 \, dx \; \text{ and } \; v=\frac 13 \sin (3x)\\ &= -\frac 13 x^2\cos (3x) + \frac 13\left [\frac 23x\sin (3x) - \int \frac 13 \sin (3x)\cdot 2 \; dx\right ]\\ &= -\frac 13 x^2\cos (3x) + \frac 29x\sin (3x) - \frac 29 \int \sin (3x) \; dx\\ &= -\frac 13 x^2\cos (3x) + \frac 29x\sin (3x) + \frac{2}{27} \cos (3x) + C. \end{align*}

(problem 3a) Compute the integral: $\int x^2\cos (2x) \;dx.$ Let $u = \answer {x^2}\;\;$ and $\;\;dv = \answer {\cos (2x) dx}$ .
Then $du = \answer {2x dx}\;\;$ and $\;\;v = \answer {\frac 12 \sin (2x)}$ .
Thus $\int x^2 \cos (2x) dx = \frac 12 x^2 \sin (2x) -\int \answer {x \sin (2x)} dx.$ Let $u = \answer {x}\;\;$ and $\;\;dv = \answer {\sin (2x) dx}$ .
Then $du = \answer {dx}\;\;$ and $\;\;v = \answer {-\frac 12 \cos (2x)}$ .
Thus $\int x^2 \cos (2x) dx = \answer {\frac 12 x^2 \sin (2x) + \frac 12 x \cos (2x) - \frac 14 \sin (2x)} + C.$

(problem 3b) Compute the integral: $\int x^2e^{-x} \;dx.$ Let $u = \answer {x^2}\;\;$ and $\;\;dv = \answer {e^{-x} dx}$ .
Then $du = \answer {2x dx}\;\;$ and $\;\;v = \answer {-e^{-x}}$ .
Thus $\int x^2 e^{-x} dx = -x^2 e^{-x} -\int \answer {-2xe^{-x}} dx.$ Let $u = 2\answer {x}\;\;$ and $\;\;dv = \answer {e^{-x} dx}$ .
Then $du = \answer {2 dx}\;\;$ and $\;\;v = \answer {-e^{-x}}$ .
Thus $\int x^2 e^{-x} dx = \answer {-x^2 e^{-x} - 2xe^{-x} - 2e^{-x}} + C.$

3 Tabular Integration

Integrals of the form $\int x^n f(x) \,dx$ require IBP $n$ times when $f(x)$ is a sine, cosine or exponential function. A notational shortcut called tabular integration can be used to streamline the process of using IBP multiple times. As an example, let’s revisit the previous example:

example 4 Compute the integral using tabular integration: $\int x^2\sin (3x) \;dx.$

We will construct a table that lists the derivatives of $u$ and the anti-derivatives of $dv$ . We can then obtain the final answer by a series of multiplications and additions or subtractions, as indicated by the arrows in the table.

Thus, $\int x^2\sin (3x) \;dx = -\frac 13 x^2\cos (3x) + \frac 29x\sin (3x) + \frac {2}{27} \cos (3x) + C.$

(problem 4) Compute the integral using Tabular Integration: $\int x^2 e^{-4x} \;dx.$ Complete the table and then enter the final answer.

$\begin {array}{c|c} u & dv \\\hline x^2 & e^{-4x} \\ \answer {2x} & \answer {-\frac 14 e^{-4x}} \\ \answer {2} & \answer {\frac {1}{16} e^{-4x}} \\ 0 & \answer {-\frac {1}{64} e^{-4x}} \end {array}$

Thus, $\displaystyle { \int x^2 e^{-4x} \;dx = }$

$-\tfrac {1}{32} (8x^2 - 4x + 1)e^{-4x} +C$ $-\tfrac {1}{32} (8x^2 + 4x + 1)e^{-4x} +C$ $\tfrac {1}{32} (8x^2 - 4x - 1)e^{-4x} +C$

example 5 Compute the integral using Tabular Integration: $\int x^4 e^{-2x} \;dx.$ We construct a table with the derivatives of $u = x^4$ and the anti-derivatives of $dv = e^{-2x} \, dx$ . We add arrows which indicate multiplication of an element of the $u$ column with elements in the $dv$ column one row lower. We adorn the arrows with alternating $+$ and $-$ signs to indicate either addition or subtraction of terms.

Thus, $\int x^4 e^{-2x} \;dx = -\frac 12 x^4 e^{-2x} - \frac 44x^3 e^{-2x} - \frac {12}{8} x^2 e^{-2x} - \frac {24}{16}x e^{-2x} - \frac {24}{32} e^{-2x} + C.$ $= -\tfrac 14\left (2 x^4 + 4x^3 + 6 x^2 + 6x + 3\right ) e^{-2x} + C.$

(problem 5) Compute the integral using Tabular Integration: $\int x^3 \cos (5x) \;dx.$ Complete the table and then enter the final answer.

$\begin {array}{c|c} u & dv \\\hline x^3 & \cos (5x) \\ \answer {3x^2} & \answer {\frac 15 \sin (5x)} \\ \answer {6x} & \answer {-\frac {1}{25} \cos (5x)} \\ \answer {6} & \answer {-\frac {1}{125} \sin (5x)} \\ 0 & \answer {\frac {1}{625} \cos (5x)} \end {array}$

Thus, $\int x^3 \cos (5x) \;dx = \answer {\frac 15 x^3\sin (5x) + \frac {3}{25}x^2 \cos (5x) - \frac {6}{125}x\sin (5x) -\frac {6}{625}\cos (5x)} +C.$

4 Inverse Functions

We will now look at some examples the involving inverse functions $\ln (x), \tan ^{-1}(x), \sin ^{-1}(x)$ and $\cos ^{-1}(x)$ . We do not (yet) know anti-derivatives for these functions, so we will have to let $u$ equal the inverse function.

example 6 Compute the integral: $\int x^5\ln (x) \;dx.$ In our previous examples, we would have let $u = x^5$ and $dv = \ln (x) \, dx$ . But notice that,in this case, we do not know an anti-derivative of $\ln (x)$ , so we would be stumped in finding $v$ . Instead, we will let $u = \ln (x) \text { and } dv = x^5 \, dx$ then $du = \frac {1}{x} \, dx \text { and } v = \frac {x^6}{6}.$ The power of $x$ has increased, rather than decreased, as in previous examples, but this is offset by the fact that the derivative of the natural log has a completely different form which is easier to work with. Now, by IBP, we have,

\begin{align*} \int x^5\ln (x) \;dx &= \frac{x^6}{6}\ln (x) - \int \frac{x^6}{6}\cdot \frac{1}{x} \, dx\\ \text{Now, the integrand } & \text{ can be simplified by canceling $x$ allowing us to integrate}\\ &= \frac{x^6}{6}\ln (x) - \frac 16\int x^5 \, dx\\ &= \frac{x^6}{6}\ln (x) - \frac{x^6}{36} + C\\ \end{align*}

(problem 6a) Compute the indefinite integral: $\int x^8 \ln (x) \;dx.$