Partial Integral

The Definition of Partial Integral

Suppose that the functions $u = u\left( x \right)$ and $v = v\left( x \right)$. The product of the two functions is determined by $y = uv$. Based on the rules for the derivative of the product of the functions, the following relationship is obtained:

$\begin{array}{l} u = u\left( x \right)\\ y = uv\\ y' = u'v + uv'\\ \frac{{dy}}{{dx}} = \frac{{du}}{{dx}}v + u\frac{{dv}}{{dx}}\\ dy = vdu + udv \end{array}$

By applying the integration operation to each side of the equation, we get:

$\begin{array}{l} \int {dy} = \int {\left( {vdu + udv} \right)} \\ \Leftrightarrow y = \int {vdu + \int {udv} } \\ \Leftrightarrow uv = \int {vdu + \int {udv} } \\ \Leftrightarrow \int {udv} = uv - \int {vdu} \end{array}$

Suppose that $u\left( x \right)$ and $v\left( x \right)$ are each function in the variable $x$, then the integration of $\int {udv} $ is determined by the relation :

\[\boxed{\int {udv} = uv - \int {vdu} }\]

The above relation shows that the integration of $\int {udv} $ can become the integration of $\int {vdu} $, and vice versa. The success or failure of the integration using the partial integral formula is determined by two things as follows:

1. Selects the $dv$ portion so that $v$ can be immediately determined via the $v = \int {dv} $ relationship.
2. $\int {vdu} $ should be easier to solve than $\int {udv} $

Example

Using the partial integral formula, determine the following integrals

1. $\int {x\sin 2xdx} $
2. $\int {x\cos 3xdx} $
3. $\int {{x^2}} \sqrt {x + 6} dx$
4. $\int {\left( {{x^2} - x} \right)} \sin 3xdx$
5. $\int {{e^x}} \cos xdx$

Answer

1. $\int {x\sin 2xdx} $
$\begin{array}{l} \int {\underbrace x_u\underbrace {\sin 2xdx}_{dv}} \\ u = x \Rightarrow du = dx\\ dv = \sin 2xdx \Rightarrow v = \int {\sin 2xdx = - \frac{1}{2}} \cos 2x\\ \int {udv} = uv - \int {vdu} \\ \Leftrightarrow \int {x\sin 2xdx = x\left( { - \frac{1}{2}\cos 2x} \right)} - \int {\left( { - \frac{1}{2}\ cos 2x} \right)} dx\\ = - \frac{1}{2}x\cos 2x - \left( { - \frac{1}{4}\sin 2x} \right) + C\\ = - \frac{1}{2}x\cos 2x + \frac{1}{4}\sin 2x + C \end{array}$

So, $\int {x\sin 2xdx} = - \frac{1}{2}x\cos 2x + \frac{1}{4}\sin 2x + C$.

2. $\int {x\cos 3xdx} $
$\begin{array}{l} \int {\underbrace x_u\underbrace {\cos 3xdx}_{dv}} \\ u = x \Rightarrow du = dx\\ dv = \cos 3xdx \Rightarrow v = \int {\cos 3xdx = \frac{1}{3}} \sin 3x\\ \int {udv} = uv - \int {vdu} \\ \Leftrightarrow \int {x\cos 3xdx = x\left( {\frac{1}{3}\sin 3x} \right)} - ​​\int {\left( {\frac{1}{3}\sin 3x } \right)} dx\\ = \frac{1}{3}x\sin 3x - \left( {\frac{1}{9}\cos 3x} \right) + C\\ = \frac{1}{3}x\sin 3x - \frac{1}{9}\cos 3x + C \end{array}$

So, $\int {x\cos 3xdx} = \frac{1}{3}x\sin 3x - \frac{1}{9}\cos 3x + C$.

3. The solution for $\int {{x^2}} \sqrt {x + 6} dx$ can use the following table:

Sign	Derivatives	Integral
$\textcolor{red}{+}$	$\textcolor{red}{{x^2}}$	${\left( {x + 6} \right)^{\frac{1}{2}}}$
$\textcolor{green}{-}$	$\textcolor{green}{2x}$	$\textcolor{red}{\frac{2}{3}{\left( {x + 6} \right)^{\frac{3}{2}}}}$
$\textcolor{blue}{+}$	$\textcolor{blue}{2}$	$\textcolor{green}{\frac{4}{{15}}{\left( {x + 6} \right)^{\frac{5}{2}}}}$
$-$	$0$	$\textcolor{blue}{\frac{8}{{105}}{\left( {x + 6} \right)^{\frac{7}{2}}}}$

From the table above, we get:

$\begin{array}{l} \int {{x^2}\sqrt {x + 6} } dx &= + {x^2}.\frac{2}{3}{\left( {x + 6} \right)^{\frac {3}{2}}} - 2x.\frac{4}{{15}}{\left( {x + 6} \right)^{\frac{5}{2}}} + 2.\frac {8}{{105}}{\left( {x + 6} \right)^{\frac{7}{2}}} + C\\ &= \frac{2}{3}{x^2}{\left( {x + 6} \right)^{\frac{3}{2}}} - \frac{8}{{15}} x{\left( {x + 6} \right)^{\frac{5}{2}}} + \frac{{16}}{{105}}{\left( {x + 6} \right) ^{\frac{7}{2}}} + C\\ &= \frac{2}{3}{x^2}\left( {x + 6} \right)\sqrt {x + 6} - \frac{8}{{15}}x{\left( { x + 6} \right)^2}\sqrt {x + 6} + \frac{{16}}{{105}}{\left( {x + 6} \right)^3}\sqrt {x + 6} + C \end{array}$

So, $\int {{x^2}\sqrt {x + 6} } dx = \frac{2}{3}{x^2}\left( {x + 6} \right)\sqrt {x + 6} - \frac{8}{{15}}x{\left( {x + 6} \right)^2}\sqrt {x + 6} + \frac{{16}}{{105 }}{\left( {x + 6} \right)^3}\sqrt {x + 6} + C$.

4. The solution for $\int {\left( {{x^2} - x} \right)} \sin 3xdx$ can use the following table:

Sign	Derivatives	Integral
$\textcolor{red}{+}$	$\textcolor{red}{{x^2} - x}$	$\sin 3x$
$\textcolor{green}{-}$	$\textcolor{green}{2x - 1}$	$\textcolor{red}{ - \frac{1}{3}\cos 3x}$
$\textcolor{blue}{+}$	$\textcolor{blue}{2}$	$\textcolor{green}{ - \frac{1}{3}.\frac{1}{3}\sin 3x = - \frac{1}{9}\sin 3x}$
$-$	$0$	$\textcolor{blue}{ - \frac{1}{9}. - \frac{1}{3}\cos 3x = \frac{1}{{27}}\cos 3x}$

From the table above, we get:

$\begin{array}{l} \int {\left( {{x^2} - x} \right)} \sin 3xdx &= + \left( {{x^2} - x} \right). - \frac{1}{3}\cos 3x - \left( {2x - 1} \right). - \frac{1}{9}\sin 3x + 2.\frac{1}{{27}}\cos 3x + C\\ &= - \frac{1}{3}\left( {{x^2} - x} \right)\cos 3x + \frac{1}{9}\left( {2x - 1} \right)\ sin 3x + \frac{2}{{27}}\cos 3x + C \end{array}$

So, $\int {\left( {{x^2} - x} \right)} \sin 3xdx = - \frac{1}{3}\left( {{x^2} - x} \right)\cos 3x + \frac{1}{9}\left( {2x - 1} \right)\sin 3x + \frac{2}{{27}}\cos 3x + C$.

5. $\int {{e^x}} \cos xdx$
$\begin{array}{l} \int {{e^x}} \cos xdx\\ \int {\underbrace {{e^x}}_u} \underbrace {\cos xdx}_{dv}\\ u = {e^x} \Rightarrow du = {e^x}dx\\ dv = \cos xdx \Rightarrow v = \int {\cos xdx = \sin x} \\ \int {\underbrace {{e^x}}_u} \underbrace {\cos xdx}_{dv} = {e^x}\sin x - \int {\sin x} .{e^x}dx\ \ = {e^x}\sin x - \int {{e^x}\sin x} dx\\ = {e^x}\sin x - \left[ {{e^x}\left( { - \cos x} \right) - \int {\left( { - \cos x} \right).} { e^x}dx} \right]\\ = {e^x}\sin x + {e^x}\cos x - \int {{e^x}} \cos xdx\\ \int {{e^x}} \cos xdx = {e^x}\sin x + {e^x}\cos x - \int {{e^x}} \cos xdx\\ 2\int {{e^x}} \cos xdx = {e^x}\sin x + {e^x}\cos x\\ \int {{e^x}} \cos xdx = \frac{{{e^x}\sin x + {e^x}\cos x}}{2} + C \end{array}$

So $\int {{e^x}} \cos xdx = \frac{{{e^x}\sin x + {e^x}\cos x}}{2} + C$.