Integration by Parts

Integration by Parts

A LevelAQAEdexcelOCRAQA November 2022Edexcel November 2022OCR November 2022

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.

A Level AQA Edexcel OCR

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}

A LevelAQAEdexcelOCR

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}

A LevelAQAEdexcelOCR

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}

A LevelAQAEdexcelOCR

Example Questions

u=2x\;\;\;\dfrac{dv}{dx}=e^{3x}

 

\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}

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}
u=x^{3}\;\;\;\dfrac{dv}{dx}=\cos(2x)

 

\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}

Additional Resources

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