Integration by Parts
Integration by Parts (IBP ) is a special method for integrating products of functions. For example, the following integrals
\[\int {x\cos xdx} ,\;\;\int {{x^2}{e^x}dx} ,\;\;\int {x\ln xdx} ,\]
in which the integrand is the product of two functions can be solved using integration by parts.
This method is based on the product rule for differentiation.
Suppose that u (x )v (x )
\[{d\left( {uv} \right) = udv + vdu.}\]
Rearranging this rule, we can write
\[udv = d\left( {uv} \right) - vdu.\]
Integrating both sides with respect to x results in
\[\int {{{u}{dv}}} = uv - \int {vdu} .\]
This is the integration by parts formula . The goal when using this formula is to replace one integral (on the left) with another (on the right), which can be easier to evaluate.
The key thing in integration by parts is to choose u and dv correctly.
The acronym ILATE is good for picking u . ILATE stands for
The closer a function is to the top, the more likely that it should be used as \(u.\)
Solved Problems
Example 1.
Compute \[\int {x\sin \left( {3x - 2} \right)dx}.\]
Solution.
We use integration by parts:
\[\int {udv} = uv - \int {vdu} .\]
Let \(u = x,\) \(dv = \sin \left( {3x - 2} \right)dx.\) Then
\[v = \int {\sin \left( {3x - 2} \right)dx} = - \frac{1}{3}\cos \left( {3x - 2} \right),\;\;du = dx.\]
Hence, the integral is
\[\int {x\sin \left( {3x - 2} \right)dx} = - \frac{x}{3}\cos \left( {3x - 2} \right) - \int {\left( { - \frac{1}{3}\cos \left( {3x - 2} \right)} \right)dx} = - \frac{x}{3}\cos \left( {3x - 2} \right) + \frac{1}{3}\int {\cos \left( {3x - 2} \right)dx} = - \frac{x}{3}\cos \left( {3x - 2} \right) + \frac{1}{3} \cdot \frac{1}{3}\sin\left( {3x - 2} \right) + C = \frac{1}{9}\sin\left( {3x - 2} \right) - \frac{x}{3}\cos \left( {3x - 2} \right) + C.\]
Example 2.
Integrate \[\int {{x}{{{\csc }^2}x}dx} \] by parts.
Solution.
We can choose \(u = x\) because \(du = dx\) is simpler. Then
\[dv = {\csc ^2}xdx = \frac{{dx}}{{{{\sin }^2}x}},\]
so we can easily integrate it and find the function \(v:\)
\[v = \int {dv} = \int {\frac{{dx}}{{{{\sin }^2}x}}} = - \cot x.\]
Apply the integration by parts formula:
\[\int {udv} = uv - \int {vdu} ,\;\; \Rightarrow \int {\underbrace x_u\underbrace {{{\csc }^2}xdx}_{dv}} = \underbrace x_u\underbrace {\left( { - \cot x} \right)}_v - \int {\underbrace {\left( { - \cot x} \right)}_v\underbrace {dx}_{du}} = - x\cot x + \int {\cot xdx} .\]
The last integral is well known:
\[\int {\cot xdx} = \ln \left| {\sin x} \right| + C.\]
Hence
\[\int {x{{\csc }^2}xdx} = - x\cot x + \ln \left| {\sin x} \right| + C.\]
Example 3.
Evaluate the integral \[\int {x\cos 2xdx}.\]
Solution.
We choose
\[u = x,\;\;dv = \cos 2xdx.\]
Hence
\[du = dv,\;\;v = \int {\cos 2xdx} = \frac{1}{2}\sin 2x.\]
Substituting these expressions into the integration by parts formula
\[\int {udv} = uv - \int {vdu} ,\]
we have
\[\int {x\cos 2xdx} = x \cdot \frac{1}{2}\sin 2x - \int {\frac{1}{2}\sin 2xdx} = \frac{x}{2}\sin 2x - \frac{1}{2}\int {\sin 2xdx} = \frac{x}{2}\sin 2x - \frac{1}{2} \cdot \left( { - \frac{1}{2}\cos 2x} \right) + C = \frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x + C.\]
Example 4.
Integrate \[\int {\ln xdx}.\]
Solution.
We are to integrate by parts: \(u = \ln x,\) \(dv = dx.\) The only choices we have for \(u\) and \(dv\) are \(du = {\frac{1}{x}} dx,\) \(v = \int {dx} = x.\) Then
\[\int {\ln xdx} = x\ln x - \int {x \cdot \frac{1}{x}dx} = x\ln x - x + C.\]
Example 5.
Evaluate the integral \[\int {\frac{{\ln x}}{{{x^2}}} dx}.\]
Solution.
Using the ILATE rule, we can choose
\[u = \ln x,\;\;dv = \frac{{dx}}{{{x^2}}}.\]
Then
\[du = \frac{{dx}}{x},\;\;v = \int {\frac{{dx}}{{{x^2}}}} = - \frac{1}{x}.\]
Integrating by parts, we obtain
\[\int {\frac{{\ln x}}{{{x^2}}}dx} = \ln x \cdot \left( { - \frac{1}{x}} \right) - \int {\left( { - \frac{1}{x}} \right)\frac{{dx}}{x}} = - \frac{{\ln x}}{x} + \int {\frac{{dx}}{{{x^2}}}} = - \frac{{\ln x}}{x} - \frac{1}{x} + C.\]
Example 6.
Evaluate the integral \[\int {{{\log }_2}xdx}.\]
Solution.
To use integration by parts we rewrite the integral as follows:
\[\int {{{\log }_2}xdx} = \int {1 \cdot {{\log }_2}xdx} \]
Now we can apply the ILATE rule, that is
\[u = {\log _2}x,\;\;dv = 1dx.\]
This yields
\[du = \frac{{dx}}{{x\ln 2}},\;\;v = \int {1dx} = x.\]
Integrating by parts, we have
\[\int {{{\log }_2}xdx} = x{\log _2}x - \int {x \cdot \frac{{dx}}{{x\ln 2}}} = x{\log _2}x - \frac{1}{{\ln 2}}\int {dx} = x{\log _2}x - \frac{x}{{\ln 2}} + C.\]
Example 7.
Compute the integral \[\int {x{2^x}dx}.\]
Solution.
Keeping in mind the ILATE rule, we can choose
\[u = x,\;\;dv = {2^x}dx.\]
Then
\[du = dx,\;\;v = \int {{2^x}dx} = \frac{{{2^x}}}{{\ln 2}}.\]
This yields:
\[\int {x{2^x}dx} = \frac{{x{2^x}}}{{\ln 2}} - \int {\frac{{{2^x}}}{{\ln 2}}dx} = \frac{{x{2^x}}}{{\ln 2}} - \frac{1}{{\ln 2}}\int {{2^x}dx} = \frac{{x{2^x}}}{{\ln 2}} - \frac{1}{{\ln 2}} \cdot \frac{{{2^x}}}{{\ln 2}} + C = \frac{{x{2^x}}}{{\ln 2}} - \frac{{{2^x}}}{{{{\left( {\ln 2} \right)}^2}}} + C = \frac{{{2^x}}}{{\ln 2}}\left( {x - \frac{1}{{\ln 2}}} \right) + C.\]
Example 8.
Evaluate the integral \[\int {x{e^{ - x}}dx}.\]
Solution.
We set
\[u = x,\;\;dv = {e^{ - x}}dx.\]
Then
\[du = dx,\;\;v = \int {{e^{ - x}}dx} = - {e^{ - x}}.\]
Using the integration by parts formula
\[\int {udv} = uv - \int {vdu} ,\]
we have
\[\int {x{e^{ - x}}dx} = x\left( { - {e^{ - x}}} \right) - \int {\left( { - {e^{ - x}}} \right)dx} = - x{e^{ - x}} + \int {{e^{ - x}}dx} = - x{e^{ - x}} - {e^{ - x}} + C = - {e^{ - x}}\left( {x + 1} \right) + C.\]
See more problems on Page 2.
