# Calculus

Integration of Functions# Integration by Parts

Problems 1-2

Problems 3-5

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

Integrating both sides of the expression, we obtain

or, rearranging terms,

Integration by using this formula is called integration by parts.

## Solved Problems

Click on problem description to see 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.}

\]

Problems 1-2

Problems 3-5