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Calculus Single Variable

Wai Yan Pong

Section 3.1 Integration by Parts

Integration by parts is the counterpart to the product rule for differentiation, applied to antiderivatives.
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Recall the product rule in differential form:
\begin{equation*} d(uv) = u\,dv + v\,du. \end{equation*}
Integrating both sides yields the following relationship between two integrals:
\begin{equation} \int u\,dv = uv - \int v\,du.\tag{3.1} \end{equation}
This formula is useful for evaluating the integral \(\int u\,dv\) whenever the integral \(\int v\,du\) is easier to compute.
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By making the variable explicit, Equation (3.1) reads:
\begin{equation} \int u(x)v'(x)\,dx = u(x)v(x) - \int v(x)u'(x)\,dx.\tag{3.2} \end{equation}
When integration by parts can be applied (i.e., \(u(x)\) and \(v(x)\) are differentiable functions), it follows from the Fundamental Theorem of Calculus that the analogous relation holds for definite integrals:
\begin{equation*} \int_a^b u(x)v'(x)\,dx = \left. u(x)v(x) \right\|_a^b - \int_a^b v(x)u'(x)\,dx. \end{equation*}
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Example 3.1.

Let us compute the indefinite integral \(\int \ln(x)\,dx\) using integration by parts (IBP). Comparing \(\ln(x)\,dx\) with \(u\,dv\text{,}\) we set \(u=\ln(x)\) and \(dv = dx\text{.}\) Then \(du = \frac{1}{x}dx\) and we can choose \(v(x)=x\text{.}\) According to Equation (3.1):
\begin{align*} \int \ln(x)\,dx &= x\ln(x) - \int x\,d(\ln(x))\\ &= x\ln(x) - \int x \cdot \frac{1}{x}\,dx\\ &= x\ln(x) - \int dx\\ &= x\ln(x) - x + C. \end{align*}
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Checkpoint 3.2.

Find the indefinite integral \(\ds \int x\cos(x)\,dx\text{.}\)
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Solution.
Set \(u=x\text{,}\) so \(du=dx\text{.}\) Let \(dv = \cos(x)\,dx\text{,}\) which implies \(v = \sin(x)\text{.}\) Applying IBP (3.1) yields:
\begin{align*} \int x\cos(x)\,dx &= x\sin(x) - \int \sin(x)\,dx\\ &= x\sin(x) - (-\cos(x)) + C\\ &= x\sin(x) + \cos(x) + C. \end{align*}
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Checkpoint 3.3.

Find the indefinite integral \(\ds \int x^2\sin(x)\,dx\text{.}\)
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Hint.
Set \(u=x^2\) and \(dv = \sin(x)\,dx\text{.}\) Then apply IBP twice.
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Example 3.4.

As another example, let us compute \(\int e^x \cos(x)\,dx\text{.}\) Let \(u = e^x\) and \(dv = \cos(x)\,dx\text{.}\) Then \(du = e^x\,dx\) and \(v = \sin(x)\text{.}\) Applying IBP yields:
\begin{equation*} \int e^x \cos(x)\,dx = e^x\sin(x) - \int e^x \sin(x)\,dx. \end{equation*}
At first glance, it seems we have made no progress since \(\int e^x \sin(x)\,dx\) appears as difficult as the original problem. However, applying IBP again to this new integral with \(u=e^x\) and \(dv = \sin(x)\,dx\) (so \(v = -\cos(x)\)), we get:
\begin{equation*} \int e^x \sin(x)\,dx = -e^x \cos(x) + \int e^x \cos(x)\,dx. \end{equation*}
For simplicity, let \(I_c = \int e^x \cos(x)\,dx\) and \(I_s = \int e^x \sin(x)\,dx\text{.}\) The equations above become:
\begin{equation*} \begin{cases} I_c &= e^x \sin(x) - I_s \\ I_s &= -e^x \cos(x) + I_c \end{cases} \end{equation*}
Substituting the second equation into the first:
\begin{align*} I_c &= e^x \sin(x) - (-e^x \cos(x) + I_c)\\ I_c &= e^x \sin(x) + e^x \cos(x) - I_c\\ 2I_c &= e^x(\sin(x) + \cos(x)) \end{align*}
Solving for \(I_c\) (and similarly for \(I_s\)), we find:
\begin{equation*} I_c = \frac{e^x}{2}(\sin(x) + \cos(x)) + C, \quad I_s = \frac{e^x}{2}(\sin(x) - \cos(x)) + C. \end{equation*}
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Checkpoint 3.5.

Compute \(\int e^x\cos(x)\,dx\) again using IBP, but this time start with \(u=\cos(x)\) and \(dv = e^x\,dx\text{.}\) Then, in the second application of IBP, use \(u=\sin(x)\) and \(dv = e^x\,dx\text{.}\)
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What happens if we use \(u = e^x\text{,}\) \(dv = \cos(x)\,dx\) in the first application of IBP and then switch to \(u=\sin(x)\text{,}\) \(dv = e^x\,dx\) in the second application?
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In some situations, IBP leads to relations called reduction formulas. For instance, using IBP, we can derive:
\begin{align*} \int \sin^n(x)\,dx &= -\sin^{n-1}(x)\cos(x) + (n-1)\int \sin^{n-2}(x)\cos^2(x)\,dx\\ &= -\sin^{n-1}(x)\cos(x) + (n-1)\int \sin^{n-2}(x)(1-\sin^2(x))\,dx\\ &= -\sin^{n-1}(x)\cos(x) + (n-1)\int \sin^{n-2}(x)\,dx - (n-1)\int \sin^n(x)\,dx. \end{align*}
Let \(S_k = \int \sin^k(x)\,dx\text{.}\) Rearranging the terms gives:
\begin{equation} S_n = -\frac{1}{n}\sin^{n-1}(x)\cos(x) + \frac{n-1}{n}S_{n-2}.\tag{3.3} \end{equation}
Each application of (3.3) reduces the power of the integrand by 2. Eventually, the problem reduces to finding either \(I_1=\int \sin(x)\,dx\) (when \(n\) is odd) or \(I_0=\int dx\) (when \(n\) is even), both of which are elementary.
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Example 3.6.

As an example, let us compute the integral \(\int \sin^3(x)\,dx\text{.}\) Using the reduction formula with \(n=3\text{:}\)
\begin{align*} \int \sin^3(x)\,dx &= S_3 = -\frac{1}{3}\sin^2(x)\cos(x) + \frac{2}{3}S_1\\ &= -\frac{1}{3}\sin^2(x)\cos(x) + \frac{2}{3}\int \sin(x)\,dx\\ &= -\frac{1}{3}\sin^2(x)\cos(x) - \frac{2}{3}\cos(x) + C. \end{align*}
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See Exercise 3.5.3 for reduction formulas involving powers of other trigonometric functions.
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Example 3.7.

Example 3.1 naturally leads to the following reduction formula for the integral of powers of \(\ln(x)\text{.}\) Let \(L_n = \int (\ln(x))^n\,dx\text{.}\) Then:
\begin{align*} L_n &= x(\ln(x))^n - \int x\,d((\ln(x))^n)\\ &= x(\ln(x))^n - \int x \cdot n(\ln(x))^{n-1} \cdot \frac{1}{x}\,dx\\ &= x(\ln(x))^n - nL_{n-1}. \end{align*}
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For instance, applying this reduction formula for \(n=3\text{:}\)
\begin{align*} L_3 &= x(\ln(x))^3 - 3L_2\\ &= x(\ln(x))^3 - 3(x(\ln(x))^2 - 2L_1)\\ &= x(\ln(x))^3 - 3x(\ln(x))^2 + 6L_1\\ &= x(\ln(x))^3 - 3x(\ln(x))^2 + 6(x\ln(x) - x) + C\\ &= x(\ln(x))^3 - 3x(\ln(x))^2 + 6x\ln(x) - 6x + C. \end{align*}
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Example 3.8.

Let us find the integral \(\ds \int x\arctan(2x)\,dx\) using integration by parts. Since the antiderivative of \(\arctan(2x)\) is not readily available, we choose \(u=\arctan(2x)\) and \(dv = x\,dx\text{.}\)
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Then \(du = \frac{2}{1+4x^2}\,dx\text{,}\) and we can take \(v=x^2/2\text{.}\) Applying IBP:
\begin{align*} \int x\arctan(2x)\,dx &= \frac{x^2\arctan(2x)}{2} - \int \frac{x^2}{1+4x^2}\,dx. \end{align*}
To integrate the rational function \(\frac{x^2}{1+4x^2}\text{,}\) we perform algebraic manipulation:
\begin{equation*} \frac{x^2}{1+4x^2} = \frac{1}{4} \cdot \frac{4x^2}{1+4x^2} = \frac{1}{4} \left(\frac{1+4x^2-1}{1+4x^2}\right) = \frac{1}{4} \left(1 - \frac{1}{1+4x^2}\right). \end{equation*}
Thus,
\begin{equation*} \int \frac{x^2}{1+4x^2}\,dx = \frac{1}{4}\left(x - \frac{1}{2}\arctan(2x)\right). \end{equation*}
Therefore, the original indefinite integral is:
\begin{gather*} \frac{x^2\arctan(2x)}{2} - \left( \frac{x}{4} - \frac{1}{8}\arctan(2x) \right) + C\\ = \frac{x^2\arctan(2x)}{2} - \frac{x}{4} + \frac{1}{8}\arctan(2x) + C. \end{gather*}
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