Integration by parts

Integration by parts (IBP) is a method of integration with the formula:

:${\begin{aligned}\int _{a}^{b}u(x)v'(x)dx&={\Big [}u(x)v(x){\Big ]}_{a}^{b}-\int _{a}^{b}u'(x)v(x)dx\\[6pt]&=u(b)v(b)-u(a)v(a)-\int _{a}^{b}u'(x)v(x)dx.\end{aligned}}$

Or more compactly,

$\int _{a}^{b}udv=uv|_{a}^{b}-\int _{a}^{b}vdu$       or without bounds       $\int udv=uv-\int vdu$

where $u$ and $v$ are functions of a variable, for instance, $x$, giving $u(x)$ and $v(x)$.

${\frac {d}{dx}}u(x)\rightarrow {\frac {du}{dx}}=u'(x)\rightarrow u'(x)dx=du$       and       $\int dv=\int v(x)=v$

Note: $dv$ is whatever terms are not included as $u$.

ILATE Rule

A rule of thumb has been proposed, consisting of choosing $u$ as the function that comes first in the following list:

I – inverse trigonometric functions: $\arctan(x),\ \operatorname {arcsec}(x),$ etc.

L – logarithmic functions: $\ln(x),\ \log _{b}(x),$ etc.

A – polynomials: $x^{2},\ 3x^{50},$ etc.

T – trigonometric functions: $\sin(x),\ \tan(x),$ etc.

E – exponential functions: $e^{x},\ 19^{x},$ etc.

Derivation

The theorem can be derived as follows. For two continuously differentiable functions $u(x)$ and $v(x)$, the product rule states:

${\Big (}u(x)v(x){\Big )}'\ =\ v(x)u'(x)+u(x)v'(x).$

Integrating both sides with respect to $x$,

$\int {\Big (}u(x)v(x){\Big )}'\,dx\ =\ \int u'(x)v(x)\,dx+\int u(x)v'(x)\,dx,$

and noting that an indefinite integral is an antiderivative gives

$u(x)v(x)\ =\ \int u'(x)v(x)\,dx+\int u(x)v'(x)\,dx,$

where we neglect writing the constant of integration. This yields the formula for integration by parts:

$\int u(x)v'(x)\,dx\ =\ u(x)v(x)-\int u'(x)v(x)\,dx,$

or in terms of the differentials $\ du=u'(x)\,dx,\ \ dv=v'(x)\,dx,\quad$

$\int u(x)\,dv\ =\ u(x)v(x)-\int v(x)\,du.$

This is to be understood as an equality of functions with an unspecified constant added to each side. Taking the difference of each side between two values x = a and x = b and applying the fundamental theorem of calculus gives the definite integral version:

$\int _{a}^{b}u(x)v'(x)\,dx\ =\ u(b)v(b)-u(a)v(a)-\int _{a}^{b}u'(x)v(x)\,dx.$

Examples

Functions multiplied by one and itself

Given $I=\int \ln(x)\cdot 1\ dx\$

The first example is ∫ ln(x) dx. We write this as:

$I=\int \ln(x)\cdot 1\ dx\ .$

Let:

$u=\ln(x)\ \Rightarrow \ du={\frac {dx}{x}}$

$dv=dx\ \Rightarrow \ v=x$

then:

${\begin{aligned}\int \ln(x)\ dx&=x\ln(x)-\int {\frac {x}{x}}\ dx\\&=x\ln(x)-\int 1\ dx\\&=x\ln(x)-x+C\end{aligned}}$

where C is the constant of integration.

Given $I=\int \arctan(x)\ dx.$

The second example is the inverse tangent function arctan(x):

$I=\int \arctan(x)\ dx.$

Rewrite this as

$\int \arctan(x)\cdot 1\ dx.$

Now let:

$u=\arctan(x)\ \Rightarrow \ du={\frac {dx}{1+x^{2}}}$

$dv=dx\ \Rightarrow \ v=x$

then

${\begin{aligned}\int \arctan(x)\ dx&=x\arctan(x)-\int {\frac {x}{1+x^{2}}}\ dx\\[8pt]&=x\arctan(x)-{\frac {\ln(1+x^{2})}{2}}+C\end{aligned}}$

using a combination of the inverse chain rule method and the natural logarithm integral condition.

Polynomials and trigonometric functions

In order to calculate

$I=\int x\cos(x)\ dx\ ,$

let:

$u=x\ \Rightarrow \ du=dx$

$dv=\cos(x)\ dx\ \Rightarrow \ v=\int \cos(x)\ dx=\sin(x)$

then:

${\begin{aligned}\int x\cos(x)\ dx&=\int u\ dv\\&=u\cdot v-\int v\,du\\&=x\sin(x)-\int \sin(x)\ dx\\&=x\sin(x)+\cos(x)+C,\end{aligned}}$

where C is a constant of integration.

For higher powers of x in the form

$\int x^{n}e^{x}\ dx,\ \int x^{n}\sin(x)\ dx,\ \int x^{n}\cos(x)\ dx\ ,$

repeatedly using integration by parts can evaluate integrals such as these; each application of the theorem lowers the power of x by one.

Exception to LIATE

$\int x^{3}e^{x^{2}}\,dx,$

one would set

$u=x^{2},\quad dv=x\cdot e^{x^{2}}\,dx,$

so that

$du=2x\,dx,\quad v={\frac {e^{x^{2}}}{2}}.$

Then

$\int x^{3}e^{x^{2}}\,dx=\int (x^{2})\left(xe^{x^{2}}\right)\,dx=\int u\,dv=uv-\int v\,du={\frac {x^{2}e^{x^{2}}}{2}}-\int xe^{x^{2}}\,dx.$

Finally, this results in

$\int x^{3}e^{x^{2}}\,dx={\frac {e^{x^{2}}(x^{2}-1)}{2}}+C.$

Performing IBP twice

$I=\int e^{x}\cos(x)\ dx.$

Here, integration by parts is performed twice. First let

$u=\cos(x)\ \Rightarrow \ du=-\sin(x)\ dx$

$dv=e^{x}\ dx\ \Rightarrow \ v=\int e^{x}\ dx=e^{x}$

then:

$\int e^{x}\cos(x)\ dx=e^{x}\cos(x)+\int e^{x}\sin(x)\ dx.$

Now, to evaluate the remaining integral, we use integration by parts again, with:

$u=\sin(x)\ \Rightarrow \ du=\cos(x)\ dx$

$dv=e^{x}\ dx\ \Rightarrow \ v=\int e^{x}\ dx=e^{x}.$

Then:

$\int e^{x}\sin(x)\ dx=e^{x}\sin(x)-\int e^{x}\cos(x)\ dx.$

Putting these together,

$\int e^{x}\cos(x)\ dx=e^{x}\cos(x)+e^{x}\sin(x)-\int e^{x}\cos(x)\ dx.$

The same integral shows up on both sides of this equation. The integral can simply be added to both sides to get

$2\int e^{x}\cos(x)\ dx=e^{x}{\bigl [}\sin(x)+\cos(x){\bigr ]}+C,$

which rearranges to

$\int e^{x}\cos(x)\ dx={\frac {1}{2}}e^{x}{\bigl [}\sin(x)+\cos(x){\bigr ]}+C'$

Problem Set

1) $\int 2x\cos(5-13x)dx$

2) $\int x\cos(x)dx$

3) $\int x^{3}\sin(3x)dx$

4) $\int e^{2x}\cos(3x)dx$

5) $\int e^{-x}(2x^{2}-3x+5)dx$

6) $\int 4\tan ^{-1}({3 \over x})dx$

7) $\int x\tan(x-4x^{2})dx$

8) $\int x^{5}\cos ^{-1}(3x)dx$

9) $\int e^{2x}3x^{-2}dx$

10) $\int x\ln(3x)dx$