Let \(u\left( x \right)\) and \(v\left( x \right)\) be differentiable functions. By the product rule,

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

Integrating both sides of the expression, we obtain

\[{uv = \int {udv} }+{ \int {vdu} ,}\]

or, rearranging terms,

\[{\int {udv} }={ uv – \int {vdu} .}\]

Integration by using this formula is called integration by parts.

## Solved Problems

Click a problem to see the solution.

### Example 1

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

Integrate \(\int {\ln xdx}.\)### Example 3

Calculate the integral \(\int {\arcsin xdx}.\)### Example 4

Find the integral \(\int {{e^x}\sin xdx}.\)### Example 5

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 \(\int {\ln xdx}.\)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.}

\]