# 8.2 | Integration by Parts

Integration by Parts: If $$f(x)$$ and $$g(x)$$ have continuous derivatives, then
$$\int f(x) g'(x) \; dx = f(x) g(x) – \int f'(x) g(x) \; dx$$
or, equivalently,
$$\int u \; dv = uv – \int v \; du$$
Find
$$\int x^3 e^x \; dx$$

$$\begin{array}{r c l} x^3 & _{\displaystyle \overset{+1}{\searrow}} & e^x \\ 3x^2 &_{\displaystyle \overset{-1}{\searrow}} & e^x \\ 6x & _{\displaystyle \overset{+1}{\searrow}} & e^x \\ 6 & _{\displaystyle \overset{-1}{\searrow}} & e^x \\ 0 & \overset{+1}{\rightarrow} & e^x \end{array}$$

By applying integration by parts 4 times, we obtain:

\begin{aligned} \int x^3 e^x \; dx & = x^3 e^x -3x^2 e^x + 6xe^x – 6e^x + C \end{aligned}
\begin{aligned} \int x^3 e^x \; dx & = x^3 e^x -3x^2 e^x + 6xe^x – 6e^x + C \end{aligned}
\begin{aligned} \int & x^3 e^x \; dx \\ & = x^3 e^x -3x^2 e^x + 6xe^x – 6e^x + C \end{aligned}
Find
$$\int t \ln t \; dt$$

$$\begin{array}{r c l} \ln t & _{\displaystyle \overset{+1}{\searrow}} & t \\ \frac{1}{t} & \overset{-1}{\rightarrow} & \frac{t^2}{2} \\ \end{array}$$

By applying integration by parts once, we obtain:
\begin{aligned} \int t \ln t \; dt & = t \ln|t| + \int \frac{-t}{2} \; dt \\ & = t \ln|t| – \int \frac{t}{2} \; dt \\ & = t \ln|t| – \frac{t^2}{4} + C \\ \end{aligned}

Find
$$\int e^{\theta} \sin \theta \; d \theta$$

$$\begin{array}{r c l} e^{\theta} & _{\displaystyle \overset{+1}{\searrow}} & \sin \theta \\ e^{\theta} & _{\displaystyle \overset{-1}{\searrow}} & – \cos \theta \\ e^{\theta} & \overset{+1}{\rightarrow} & – \sin \theta \\ \end{array}$$

By applying integration by parts twice, we obtain:

\begin{aligned} \int e^{\theta} \sin \theta \; d \theta & = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta + \int -e^{\theta} \sin \theta \; d\theta \\ & = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta – \int e^{\theta} \sin \theta \; d \theta \\ \end{aligned}
\begin{aligned} \int e^{\theta} \sin \theta \; d \theta & = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta + \int -e^{\theta} \sin \theta \; d\theta \\ & = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta – \int e^{\theta} \sin \theta \; d \theta \\ \end{aligned}
\begin{aligned} & \int e^{\theta} \sin \theta \; d \theta \\ & = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta + \int -e^{\theta} \sin \theta \; d\theta \\ & = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta – \int e^{\theta} \sin \theta \; d \theta \\ \end{aligned}

So,
$$2 \int e^{\theta} \sin \theta \; d \theta = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta$$
and
$$\int e^{\theta} \sin \theta \; d \theta = \frac{-e^{\theta} \cos \theta+ e^{\theta} \sin \theta}{2} + C$$

Find
$$\int (3t^2-5t+10) e^{2t} \; dt$$

$$\begin{array}{r c l} 3t^2-5t+10 & _{\displaystyle \overset{+1}{\searrow}} & e^{2t} \\ 6t-5 &_{\displaystyle \overset{-1}{\searrow}} & \frac{e^{2t}}{2} \\ 6 & _{\displaystyle \overset{+1}{\searrow}} & \frac{e^{2t}}{4} \\ 0 & _{\displaystyle \overset{-1}{\rightarrow}} & \frac{e^{2t}}{8} \\ \end{array}$$

By applying integration by parts 3 times, we obtain:

\begin{aligned} \int (3t^2-5t+10) e^{2t} \; dt & = \frac{3t^2-5t+10}{2} e^{2t} -\frac{6t-5}{4} e^{2t} + \frac{6}{8}e^{2t} + C\\ & = \frac{3t^2-5t+10}{2} e^{2t} -\frac{6t-5}{4} e^{2t} + \frac{3}{4}e^{2t} + C \end{aligned}
\begin{aligned} & \int (3t^2-5t+10) e^{2t} \; dt \\ & = \frac{3t^2-5t+10}{2} e^{2t} -\frac{6t-5}{4} e^{2t} \\ & \; \; + \frac{6}{8}e^{2t} + C\\ & = e^{2t} \left( \frac{3t^2-5t+10}{2} -\frac{6t-5}{4} + \frac{3}{4}\right) + C \end{aligned}
\begin{aligned} & \int (3t^2-5t+10) e^{2t} \; dt \\ & = \frac{3t^2-5t+10}{2} e^{2t} -\frac{6t-5}{4} e^{2t} \\ & \; \; + \frac{6}{8}e^{2t} + C\\ & = e^{2t} \left( \frac{3t^2-5t+10}{2} -\frac{6t-5}{4} + \frac{3}{4}\right) + C \end{aligned}