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 $u$ .

LIATE Rule

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

L – logarithmic functions:

\ln(x),\ \log _{b}(x),

etc.

I – inverse trigonometric functions:

\arctan(x),\ \operatorname {arcsec}(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'

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LIATE Rule

Derivation

Examples

Functions multiplied by one and itself

Given I = ∫ ln ⁡ ( x ) ⋅ 1 d x {\displaystyle I=\int \ln(x)\cdot 1\ dx\ }

Given I = ∫ arctan ⁡ ( x ) d x . {\displaystyle I=\int \arctan(x)\ dx.}

Polynomials and trigonometric functions

Exception to LIATE

Performing IBP twice

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

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