# Integration by parts Integrals like \begin{equation} \int x e^{-6 x} \d{x} \end{equation} are not solvable by [substitution](integration-substitution.md), but they look like they are related closely enough that you may be able to! What to do here? Recall the [product rule](derivative-rules.md) for two functions *u* and *v*, then integrate: \begin{align} (uv)' &= uv' + v u' \\ uv &= \int u \d{v} + \int v \d{u} \end{align} Rearranging gives the familiar form: ```{topic} Integration by parts \begin{equation} \int u \d{v} = uv - \int v \d{u} \end{equation} To apply this technique, identify a *u* and a $\d{v}$. Differentiate $u$ and integrate $\d{v}$, then apply the formula. ``` Let's try this for the integral above with \begin{align} u &= x & \d{v} &= e^{-6x} \d{x} \\ \d{u} &= \d{x} & v &= -\frac{e^{-6x}}{6} \end{align} so \begin{align} \int x e^{-6x} \d{x} &= x\frac{-e^{-6x}}{6} - \int\frac{-1}{6} e^{-6x} \d{x} \\ &= -\frac{xe^{-6x}}{6} + \frac{1}{6} \int e^{-6x} \d{x} \\ &= -\frac{xe^{-6x}}{6} - \frac{e^{-6x}}{36} + c \end{align} ## Choosing parts The heart of the technique is making a suitable choice for *u* and $\d{v}$. The acronym L.I.A.T.E. can help make this choice: - **L**ogarithmic (best choice of *u*) - **I**nverse - **A**lgebraic - **T**rigonometric - **E**xponential (worst choice of *u*) ```{example} Integration by parts 1 Evaluate \begin{equation} \int(3x+5)\cos\left(\frac{x}{4}\right) \d{x} \end{equation} --- Let \begin{align} u &= 3x+5 & \d{v} &= \cos\left(\frac{x}{4}\right) \d{x} \\ \d{u} &= 3 \d{x} & v &= 4\sin\left(\frac{x}{4}\right) \end{align} so: \begin{align} \int(3x+5)\cos\left(\frac{x}{4}\right) \d{x} &= (3x+5)\left[4\sin\left(\frac{x}{4}\right)\right] - \int 4 \sin \left(\frac{x}{4}\right) 3 \d{x} \\ &= (12x+20) \sin\left(\frac{x}{4}\right) - 12\int\sin\left(\frac{x}{4}\right) \d{x} \\ &= (12x+20) \sin\left(\frac{x}{4}\right) + 48\cos\left(\frac{x}{4}\right) + c \end{align} ``` ```{example} Integration by parts 2 Evaluate \begin{equation} \int \ln x \d{x} \end{equation} --- Let \begin{align} u &= \ln x & \d{v} &= \d{x} \\ \d{u} &= \frac{1}{x} \d{x} & v &= x \end{align} so: \begin{align} \int \ln x \d{x} &= x \ln x - \int\frac{1}{x} x \d{x} \\ &= x \ln x - \int \d{x} \\ &= x \ln x - x + c \\ \end{align} ``` ```{example} Integration by parts 3 Evaluate \begin{equation} \int x^5 \sqrt{x^3+1} \d{x} \end{equation} --- Let \begin{align} u &= x^3 & \d{v} &= x^2\sqrt{x^3+1} \d{x} \\ \d{u} &= 3x^2 \d{x} & v &= \frac{2}{9}(x^3+1)^{3/2} \end{align} so: \begin{align} \int(x^5)(\sqrt(x^3+1))dx &= x^3\frac{2}{9}(x^3+1)^{3/2} - \int\left[\frac{2}{9}(x^3+1)^{3/2}\right]3x^2dx \\ &= \frac{2}{9}x^3(x^3+1)^{3/2} - \frac{2}{9}\cdot\frac{2}{5}(x^3+1)^{5/2} + c \\ &= \frac{2}{9}x^3(x^3+1)^{3/2} - \frac{4}{45}(x^3+1)^{5/2} + c \end{align} ``` ## Tabular method To integrate \begin{equation} \int x^4 e^{x/2} \d{x} \end{equation} by parts, let \begin{align} u &= x^4 & \d{v} &= e^{x/2} \\ \d{u} &= 4x^3 \d{x} & v &= 2e^{x/2} \end{align} which allows: \begin{equation} \int x^4e^{x/2} \d{x} = 2 x^4 e^{x/2} - \int 8 x^3 e^{x/2} \d{x} \end{equation} We will need to integrate by parts again, and again, and again... but there is a shortcut! Make a staggered table. Differentiate *u* repeatedly until you get a 0 in the last row. Then, integrate $\d{v}$ repeatedly until you reach the last nonzero row for *u*. Multiply across rows using opposite signs. | sign | $u$ | $\d{v}$ | |------|---------|-------------| | | | $e^{x/2}$ | | $+$ | $x^4$ | $2e^{x/2}$ | | $-$ | $4x^3$ | $4e^{x/2}$ | | $+$ | $12x^2$ | $8e^{x/2}$ | | $-$ | $24x$ | $16e^{x/2}$ | | $+$ | $24$ | $32e^{x/2}$ | | | $0$ | | The result is: \begin{align} \int x^4 e^{x/2} \d{x} &= x^4 (2e^{x/2}) - 4x^3 (4e^{x/2}) \\ &+ 12x^2 (8e^{x/2}) - 24x (16e^{x/2}) + 24 (32e^{x/2}) \\ &= (2x^4 - 16x^3 + 96x^2 - 384x + 768) e^{x/2} \end{align} ## Skill builder problems Evaluate the following 1. $\displaystyle \int (x+2)\ln x \d{x}$ ```{solution} Select the following parts: \begin{align} u &= \ln x & \d{v} &= (x+2)\d{x} \\ \d{u} &= \frac{1}{x}\d{x} & v &= \frac{x^2}{2} + 2x \end{align} so \begin{align} \int (x+2)\ln x \d{x} &= \left( \frac{x^2}{2} + 2x \right)\ln x - \int \left( \frac{x^2}{2} + 2x \right) \frac{1}{x} \d{x} \\ &= \left( \frac{x^2}{2} + 2x \right)\ln x - \int \left( \frac{x}{2} + 2 \right) \d{x} \\ &= \left( \frac{x^2}{2} + 2x \right)\ln x - \frac{x^2}{4} - 2x + c \end{align} ``` 2. $\displaystyle \int x^5\sin x \d{x}$ ```{solution} Since $x^5$ is to the fifth order, integration by parts must be used 5 times. Use the tabular method for this. | sign | $u$ | $\d{v}$ | |------|---------|-----------| | | | $\sin x$ | | $+$ | $x^5$ | $-\cos x$ | | $-$ | $5x^4$ | $-\sin x$ | | $+$ | $20x^3$ | $\cos x$ | | $-$ | $60x^2$ | $\sin x$ | | $+$ | $120x$ | $-\cos x$ | | $-$ | $120$ | $-\sin x$ | | | $0$ | | so \begin{equation} \int x^5\sin x \d{x} = -x^5 \cos x + 5x^4 \sin x + 20x^3 \cos x - 60x^2 \sin x - 120x \cos x + 120 \sin x + c \end{equation} To make the solution look cleaner, you can factor out $\cos x$ and $\ sin x$. \begin{equation} \int x^5\sin x \d{x} = (-x^5 + 20x^3 - 120x) \cos x + (5x^4 - 60x^2 + 120) \sin x + c \end{equation} ``` 3. $\displaystyle \int e^x\cos x \d{x}$ ```{solution} Select the following parts: \begin{align} u &= \cos x & \d{v} &= e^x\d{x} \\ \d{u} &= -\sin x\d{x} & v &= e^x \end{align} so \begin{equation} \int e^x \cos x \d{x} = e^x \cos x + \int e^x \sin x \d{x} \end{equation} To evaluate this integral, select new parts: \begin{align} u &= \sin x, & \d{v} &= e^x\d{x} \\ \d{u} &= \cos x\d{x} & v &= e^x \end{align} so \begin{equation} \int e^x \cos x \d{x} = e^x \cos x+ \left[e^x \sin x - \int e^x \cos x \d{x}\right] \end{equation} The same integral appears on both sides! Add them together and solve: \begin{align} 2 \int e^x \cos x \d{x} &= e^x \cos x + e^x \sin x \\ \int e^x \cos x \d{x} &= \frac{e^x}{2}(\cos x + \sin x) + c \end{align} ```