Calculus

Integration of Functions

Integration of Functions Logo

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