Revise
Integration by Parts
Integration by Parts Revision
Integration by Parts
We have already seen the reverse chain rule. Integration by parts is the reverse product rule.
Integration by parts has many uses, most notably integrating things of the form x^{n}f(x).
For some questions, you need to integrate by parts more than once to get a result.
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Integration by Parts
Recall: The product rule.
\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}
Integration by parts is the reverse of this.
{\LARGE \int}u\dfrac{dv}{dx}=uv-{\LARGE \int}v\dfrac{du}{dx}
You should choose u to be something that differentiates nicely, so that you can evaluate the integral on the right-hand side.
Example: Find \int x\sin(x)dx
u=x\;\;\;\dfrac{dv}{dx}=\sin(x)
\dfrac{du}{dx}=1\;\;\;v=-\cos(x)
\begin{aligned}\int x\sin(x)dx&=-x\cos(x)-\int-\cos(x)dx\\[1.2em]&=-x\cos(x)+\int \cos(x)dx\\[1.2em]&=-x\cos(x)+\sin(x)+c\end{aligned}
Integral of the Natural Logarithm
One of the most important uses of integration by parts is that it gives us a way to find \int \ln(x)dx, by writng \ln(x) as
1\times \ln(x).
u=\ln(x)\;\;\;\dfrac{dv}{dx}=1
\dfrac{du}{dx}=\dfrac{1}{x}\;\;\;v=x
\begin{aligned}\int \ln(x)dx&=x\ln(x)-\int x\dfrac{1}{x}dx\\[1.2em]&=x\ln(x)-\int1dx\\[1.2em]&=x\ln(x)-x+c\\[1.2em]&=x(\ln(x)-1)+c\end{aligned}
Integrating by Parts Multiple Times
Some integrals require you to integrate by parts multiple times to get a solution to.
Example: Find \int x^{2}e^{x}dx
u=x^{2}\;\;\;\dfrac{dv}{dx}=e^{x}
\dfrac{du}{dx}=2x\;\;\;v=e^{x}
\begin{aligned}\int x^{2}e^{x}dx=x^{2}e^{x}-\int2xe^{x}dx\end{aligned}
We must integrate by parts again the second integral.
u=2x\;\;\;\dfrac{dv}{dx}=e^{x}
\dfrac{du}{dx}=2\;\;\;v=e^{x}
\begin{aligned}\int x^{2}e^{x}dx&=x^{2}e^{x}-2xe^{x}+\int2e^{x}dx\\[1.2em]&=x^{2}e^{x}-2xe^{x}+2e^{x}+c\end{aligned}
Integration by Parts Example Questions
Question 1: What is \int 2xe^{3x}dx?
[2 marks]
\dfrac{du}{dx}=2\;\;\;v=\dfrac{1}{3}e^{3x}
\begin{aligned}\int 2xe^{3x}dx&=\dfrac{2}{3}xe^{3x}-\int\dfrac{2}{3}e^{3x}dx\\[1.2em]&=\dfrac{2}{3}xe^{3x}-\dfrac{2}{9}e^{3x}+c\end{aligned}
Question 2: Find \int 4\ln(5x)dx
[2 marks]
Have that \int \ln(x)dx=x(\ln(x)-1)+c
The 4 will remain at the front, and the 5 inside the function becomes a \dfrac{1}{5} at the front when integrating.
\begin{aligned}\int 4\ln(5x)dx&=4\times\dfrac{1}{5}\times 5x(\ln(5x)-1)+c\\[1.2em]&=4x(\ln(5x)-1)+c\end{aligned}
Question 3: Find \int x^{3}\cos(2x)dx
[4 marks]
\dfrac{du}{dx}=3x^{2}\;\;\;v=\dfrac{1}{2}\sin(2x)
\begin{aligned}\int x^{3}\cos(2x)dx=\dfrac{1}{2}x^{3}\sin(2x)-\int\dfrac{3}{2}x^{2}\sin(2x)dx\end{aligned}
Integrate the integral on the right-hand side by parts.
u=\dfrac{3}{2}x^{2}\;\;\;\dfrac{dv}{dx}=\sin(2x)
\dfrac{du}{dx}=3x\;\;\;v=-\dfrac{1}{2}\cos(2x)
\begin{aligned}\int x^{3}\cos(2x)dx&=\dfrac{1}{2}x^{3}\sin(2x)+\dfrac{3}{4}x^{2}\cos(2x)-\int\dfrac{3}{2}x\cos(2x)dx\end{aligned}
Use integration by parts again.
u=\dfrac{3}{2}x\;\;\;\dfrac{dv}{dx}=\cos(2x)
\dfrac{du}{dx}=\dfrac{3}{2}\;\;\;v=\dfrac{1}{2}\sin(2x)
\begin{aligned}\int x^{3}\cos(2x)dx&=\dfrac{1}{2}x^{3}\sin(2x)+\dfrac{3}{4}x^{2}\cos(2x)-\dfrac{3}{4}x\sin(2x)+\int\dfrac{3}{4}\sin(2x)dx\\[1.2em]&=\dfrac{1}{2}x^{3}\sin(2x)+\dfrac{3}{4}x^{2}\cos(2x)-\dfrac{3}{4}x\sin(2x)-\dfrac{3}{8}\cos(2x)+c\end{aligned}
