Integration by Parts (IBP) is a special method for integrating products of functions. For example, the following integrals

\[{\int {x\cos xdx} ,\;\;}\kern0pt{\int {{x^2}{e^x}dx} ,\;\;}\kern0pt{\int {x\ln xdx} ,}\]

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 \(u\left( x \right)\) and \(v\left( x \right)\) are differentiable functions. Then the product rule in terms of differentials gives us:

\[{d\left( {uv} \right) = udv + vdu.}\]

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

I | Inverse Trigonometric Functions \(\arcsin x, \arctan x, \ldots\) |

L | Logarithmic Functions \(\ln x, \log x, {\log_2x}, \ldots\) |

A | Algebraic Functions \(x, {x^2}, {x^3}, {2x^5}, \ldots\) |

T | Trigonometric Functions \(\sin x, \cos x, \tan x, \ldots\) |

E | Exponential Functions \({e^x}, {e^{2x}}, {2^x}, {3^{-x}}, \ldots\) |

The closer a function is to the top, the more likely that it should be used as \(u.\)

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Compute \(\int {x\sin \left( {3x – 2} \right)dx}.\)### Example 2

Integrate \(\int {{x}{{{\csc }^2}x}dx} \) by parts.### Example 3

Evaluate the integral \(\int {x\cos 2xdx}.\)### Example 4

Integrate \(\int {\ln xdx}.\)### Example 5

Evaluate the integral \(\int {\large{\frac{{\ln x}}{{{x^2}}}}\normalsize dx}.\)### Example 6

Evaluate the integral \(\int {{{\log }_2}xdx}.\)### Example 7

Compute the integral \(\int {x{2^x}dx}.\)### Example 8

Evaluate the integral \(\int {x{e^{ – x}}dx}.\)### Example 9

Compute the integral \(\int {{x^2}{e^x}dx}.\)### Example 10

Calculate the integral \(\int {\arcsin xdx}.\)### Example 11

Find the integral \(\int {\arctan xdx}.\)### Example 12

Find the integral \(\int {{e^x}\sin xdx}.\)### Example 13

Find the integral \(\int {{e^{ – x}}\sin xdx}.\)### Example 14

Compute the integral \(\int {{{\sin }^2}xdx}.\)### Example 15

Compute the integral \(\int {{{\cos }^2}xdx}.\)### Example 16

Find a reduction formula for \(\int {{{\sin }^n}xdx} ,\) \(n \ge 2.\)### 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),\;\;}\kern-0.3pt

{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

\[{u = x,\;\;}\kern0pt{dv = \cos 2xdx.}\]

Hence

\[{du = dv,\;\;}\kern0pt{v = \int {\cos 2xdx} }={ \frac{1}{2}\sin 2x.}\]

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 \(du = {\large\frac{1}{x}\normalsize} dx,\) \(v = \int {dx} = x.\) Then

\[

{\int {\ln xdx} }

= {{x\ln x }-{ \int {x \cdot \frac{1}{x}dx} }}

= {x\ln x – x }+{ C.}

\]

### Example 5.

Evaluate the integral \(\int {\large{\frac{{\ln x}}{{{x^2}}}}\normalsize dx}.\)Solution.

Using the ILATE rule, we can choose

\[{u = \ln x,\;\;}\kern0pt{dv = \frac{{dx}}{{{x^2}}}.}\]

Then

\[{du = \frac{{dx}}{x},\;\;}\kern0pt{v = \int {\frac{{dx}}{{{x^2}}}} = – \frac{1}{x}.}\]

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

\[{\int {{{\log }_2}xdx} }={ x{\log _2}x – \int {x \cdot \frac{{dx}}{{x\ln 2}}} }={ x{\log _2}x – \frac{1}{{\ln 2}}\int {dx} }={ x{\log _2}x – \frac{x}{{\ln 2}} + C.}\]