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Notes

Practice Problems

\(\textbf{1)}\) \(\displaystyle \int xe^{2x} \,dx\)

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The answer is \(\displaystyle \frac{xe^{2x}}{2} \, – \, \frac{e^{2x}}{4}+C\)

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\(\,\,\,\,\,u=x, \,\,\, dv=e^{2x}\,dx\)


\(\,\,\,\,\,du=dx, \,\,\, v=\frac{1}{2}e^{2x}\)


\(\,\,\,\,\,\int u \,dv= uv\, – \int v \,du\)


\(\,\,\,\,\,\int x e^{2x}\,dx= \left(x\right)\left(\frac{1}{2}e^{2x}\right) – \int \frac{1}{2}e^{2x} \,dx\)


\(\,\,\,\,\,\int x e^{2x}\,dx= \frac{1}{2}xe^{2x} – \frac{1}{4}e^{2x}+C\)


\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{xe^{2x}}{2} \, – \, \frac{e^{2x}}{4}+C\)

\(\textbf{2)}\) \(\displaystyle \int \theta \sin{5\theta} \,d\theta\)

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The answer is \(\displaystyle -\,\frac{\theta \cos{5\theta}}{5} +\frac{\sin{5\theta}}{25} +C\)

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\(\,\,\,\,\,u=\theta, \,\,\, dv=\sin{5\theta} \,d\theta\)


\(\,\,\,\,\,du=d\theta, \,\,\, v=-\frac{1}{5}\cos{5\theta}\)


\(\,\,\,\,\,\int u \,dv= uv\, – \int v \,du\)


\(\,\,\,\,\,\int \theta \sin{5\theta} \,d\theta= \left(\theta\right)\left(-\frac{1}{5}\cos{5\theta}\right) – \int -\frac{1}{5}\cos{5\theta} \,d\theta\)


\(\,\,\,\,\,\int \theta \sin{5\theta} \,d\theta= -\,\frac{\theta \cos{5\theta}}{5} +\frac{1}{5} \int \cos{5\theta} \,d\theta\)


\(\,\,\,\,\,\int \theta \sin{5\theta} \,d\theta= -\,\frac{\theta \cos{5\theta}}{5} +\frac{1}{5} \cdot \frac{1}{5} \sin{5\theta}+C\)


\(\,\,\,\,\,\)The answer is \(\displaystyle -\,\frac{\theta \cos{5\theta}}{5} +\frac{\sin{5\theta}}{25} +C\)

\(\textbf{3)}\) \(\displaystyle \int x \ln{x} \,dx\)

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The answer is \(\displaystyle \frac{x^2\ln{x}}{2}-\,\frac{x^2}{4}+C\)

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\(\,\,\,\,\,u=\ln{x}, \,\,\, dv= x \,dx\)


\(\,\,\,\,\,du=\frac{1}{x}\,dx, \,\,\, v=\frac{x^2}{2}\)


\(\,\,\,\,\,\int u \,dv= uv\, – \int v \,du\)


\(\,\,\,\,\,\int x \ln{x} \,dx= \frac{x^2}{2} \ln{x} – \frac{1}{2} \int x \,dx\)


\(\,\,\,\,\,\int x \ln{x} \,dx= \frac{x^2}{2} \ln{x} – \frac{1}{2} \cdot \frac{x^2}{2} + C\)


\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{x^2\ln{x}}{2}-\,\frac{x^2}{4}+C\)

\(\textbf{4)}\) \(\displaystyle \int \frac{\ln{x}}{x^6} \,dx\)

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The answer is \(-\displaystyle\frac{\ln{x}}{5x^5} – \frac{1}{25x^5} + C\)[/latex]

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\(\,\,\,\,\,u=\ln{x}, \,\,\, dv=\frac{1}{x^6} \,dx\)


\(\,\,\,\,\,du=\frac{1}{x}\,dx, \,\,\, v=\frac{x^{-5}}{-5}=\frac{-1}{5x^5}\)


\(\,\,\,\,\,\int u \,dv= uv\, – \int v \,du\)


\(\,\,\,\,\,\int \frac{\ln{x}}{x^6} \,dx= \left( \ln{x} \right) \left( -\frac{1}{5x^5} \right) – \int \left( -\frac{1}{5x^5} \right) \left( \frac{1}{x} \right) \,dx\)


\(\,\,\,\,\,\int \frac{\ln{x}}{x^6} \,dx= -\frac{\ln{x}}{5x^5} + \frac{1}{5} \int \frac{1}{x^6} \,dx\)


\(\,\,\,\,\,\int \frac{\ln{x}}{x^6} \,dx= -\frac{\ln{x}}{5x^5} + \frac{1}{5} \int x^{-6} \,dx\)


\(\,\,\,\,\,\int \frac{\ln{x}}{x^6} \,dx= -\frac{\ln{x}}{5x^5} + \frac{1}{5} \cdot \frac{x^{-5}}{-5} + C\)


\(\,\,\,\,\,\text{The answer is } -\displaystyle\frac{\ln{x}}{5x^5} – \frac{1}{25x^5} + C\)

\(\textbf{5)}\) \(\displaystyle \int e^x \cos(x) \,dx\)

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The answer is \(\frac{1}{2} e^x \sin(x) + \frac{1}{2} \cos(x) \,e^x + C\)[/latex]

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\(\,\,\,\,\,\)\(\,\,\,\,\,\) \(u = e^x, \quad dv = \cos(x) \,dx \)

\(\,\,\,\,\,\) \(du = e^x \,dx, \quad v = \int \cos(x) \,dx = \sin(x)\)

\(\,\,\,\,\,\) The integration by parts formula gives:

\(\,\,\,\,\,\) \(\int e^x \cos(x) \,dx = e^x \sin(x) – \int \sin(x) \,e^x \,dx\)

\(\,\,\,\,\,\) Now, solve the new integral \(\int \sin(x) \,e^x \,dx\) using integration by parts again:

\(\,\,\,\,\,\) Solve the integral \(\int \sin(x) \,e^x \,dx\)

\(\,\,\,\,\,\) Apply integration by parts with \(u = \sin(x)\) and \(dv = e^x \,dx\)

\(\,\,\,\,\,\) \(u = \sin(x), \quad dv = e^x \,dx \)

\(\,\,\,\,\,\) \(du = \cos(x) \,dx, \quad v = \int e^x \,dx = e^x\)

\(\,\,\,\,\,\) The integration by parts formula gives:

\(\,\,\,\,\,\) \(\int \sin(x) \,e^x \,dx = -\cos(x) \,e^x – \int (-\cos(x)) \,e^x \,dx\)

\(\,\,\,\,\,\) \(\int \sin(x) \,e^x \,dx = -\cos(x) \,e^x + \int \cos(x) \,e^x \,dx\)

\(\,\,\,\,\,\) Substitute this back into the equation for the first integral

\(\,\,\,\,\,\) \(\int e^x \cos(x) \,dx = e^x \sin(x) + \cos(x) \,e^x – \int \cos(x) \,e^x \,dx\)

\(\,\,\,\,\,\) Move the last integral term to the left side:

\(\,\,\,\,\,\) \(2 \int e^x \cos(x) \,dx = e^x \sin(x) + \cos(x) \,e^x\)

\(\,\,\,\,\,\) Solve for the original integral:

\(\,\,\,\,\,\) \(\int e^x \cos(x) \,dx = \frac{1}{2} e^x \sin(x) + \frac{1}{2} \cos(x) \,e^x + C\)

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In Summary

Integration by parts is a technique used in calculus to evaluate definite or indefinite integrals. It allows us to evaluate integrals that would otherwise be difficult or impossible to solve using other methods.

The notes for integration by parts are as follows.

\(\int u \,dv= uv \,- \int v \,du\)

In its most basic form, integration by parts states that the integral of the product of two functions can be expressed as the product of one function and the derivative of the other, minus the integral of the derivative of the first function multiplied by the second.

Integration by parts is typically taught in a calculus course, either at the high school or college level. It is often introduced in conjunction with other integration techniques, such as substitution and partial fractions.

Related topics to Integration by Parts: Some related topics to integration by parts include the fundamental theorem of calculus, differential equations, and series.

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