Calculus

Integration of Functions

Integration by Parts

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

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

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