Solution

Recall the formula for integration by parts:

\[\int_a^b uv'\;\mathrm{d} x=\Bigl[uv\Bigl]_a^b-\int_a^b u'v\;\mathrm{d} x.\]

The first step is to make suitable choices for $u$ and $v'$. The two functions appearing in the integrand are $1+x$ and $\sin{x}$. The choice for $u$ should be a function that will become simpler when differentiated; here $\sin{x}$ would be a bad choice, as the derivative of $\sin{x}$ is just another trigonometric function, $\cos{x}$. The derivative of $1+x$, however, is a constant, so we choose $u = 1+x$.

The choices for $u$ and $v'$ are $u=1+x$ and $v'=\sin{x}$.

It can be convenient to lay out the expressions for $u$, $v$ and their derivatives in the following way:

\begin{align} u &= 1+x, & v &= \; ???, \\ u' &= \; ???, & v' &= \sin x. \end{align}

The gaps for $u'$ and $v$ can be filled in after performing the relevant calculations.

Differentiating $u$ with respect to $x$ gives $u'$:

\[u' = \frac{\mathrm{d </div>{\mathrm{d} x}\bigl[1+x\bigr] = 1,\]

and integrating $v'$ with respect to $x$ gives $v$:

\[v = \int \sin{x} \mathrm{d} x = -\cos{x}.\]

Note: It is not necessary to include the constant of integration when computing $v$.

The expressions for $u$, $v$, $u'$ and $v'$ are:

\begin{align} u &= 1+x, & v &= -\cos{x}, \\ u' &= 1, & v' &= \sin x. \end{align}

Substituting these into the formula for integration by parts gives:

\begin{align} \int_0^{\pi/2} (1+x)\sin{x}\;\mathrm{d} x &= \Bigl[(1+x)(-\cos{x})\Bigl]_0^{\pi/2}-\int_0^{\pi/2} 1\cdot(-\cos{x})\;\mathrm{d} x \\ &= \Bigl[-(1+x)\cos{x}\Bigl]_0^{\pi/2}+\int_0^{\pi/2} \cos{x}\;\mathrm{d} x. \end{align}

First evaluate $\Bigl[-(1+x)\cos{x}\Bigl]_0^{\pi/2}$

\begin{align} \Bigl[-(1+x)\cos{x}\Bigl]_0^{\pi/2} &= -\Bigl(1+\dfrac{\pi}{2}\Bigl)\cos{\dfrac{\pi}{2} }-\Bigl\{-(1+0)\cos{0}\Bigl\} \\ &= -\Bigl(1+\frac{\pi}{2}\Bigl)\cdot\,0+1\cdot1 \\ &= 1. \end{align}

Substitute this back in the formula to obtain:

\begin{align}\int_0^{\pi/2} (1+x)\sin{x}\;\mathrm{d} x &= 1 + \int_0^{\pi/2} \cos{x}\;\mathrm{d} x \\ &= 1 + \Bigl[\sin{x}\Bigl]_0^{\pi/2} \\ &= 1 + \bigl(\sin{\frac{\pi}{2} }-\sin{0}\bigl) \\ &= 1+\bigl(1-0\bigl) \\ &= 2. \end{align}

The value of the integral is $\begin{align}\int_0^{\pi/2} (1+x)\sin{x}\;\mathrm{d} x=2\end{align}$.

Note: Suppose different choices were made for $u$ and $v'$. Let $u=\sin{x}$ and $v'=(1+x)$. Then $u'=\cos{x}$ and $v=x+\dfrac{x^2}{2}$. Substituting these expressions into the formula for integration by parts yields: \[\int_0^{\pi/2} (1+x)\sin{x}\;\mathrm{d} x=\left[\Bigl(x+\dfrac{x^2}{2}\Bigl)\sin{x}\right]_0^{\pi/2}-\int_0^{\pi/2} \left(x+\dfrac{x^2}{2}\right)\cos{x}\;\mathrm{d} x.\] It is immediately obvious that the integral on the right hand side is more complicated than the original problem, thus emphasising the importance of making appropriate choices for $u(x)$ and $v'(x)$.

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Video Examples

Example 1

Prof. Robin Johnson uses integration by parts to find $\begin{align}\int_0^{\large{\pi} } x^2\sin{x}\;\mathrm{d} x\end{align}$.

Example 2

Prof. Robin Johnson uses integration by parts to find $\begin{align}\int_0^{\large{\pi} }\mathrm{e}^{-t}\sin{t}\;\mathrm{d} t\end{align}$.

Example 3

Prof. Robin Johnson uses integration by parts to find $\begin{align}\int(1+2x)\mathrm{e}^{\large{-x} }\;\mathrm{d} x\end{align}$.

Example 4

Prof. Robin Johnson uses integration by parts to find $\begin{align}\int x\sin{(1-2x)}\;\mathrm{d} x\end{align}$.

Example 5

Prof. Robin Johnson uses integration by parts to find $\begin{align}\int x\arctan{x}\;\mathrm{d} x\end{align}$.

Workbook

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

• Integration by parts

Test Yourself

Test yourself: Numbas test on integration by parts

External Resources

• Integration by parts workbook at mathcentre
• Definite integration by parts workbook at mathcentre
• Integration by parts videos at Khan Academy