# Integration by Parts

• 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

${\int {\ln xdx} } = {{x\ln x }-{ \int {x \cdot \frac{1}{x}dx} }} = {x\ln x – x }+{ C.}$

Page 1
Problems 1-2
Page 2
Problems 3-5