Integration by parts

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

: ∫ a b u ( x ) v ′ ( x ) d x = [ u ( x ) v ( x ) ] a b − ∫ a b u ′ ( x ) v ( x ) d x = u ( b ) v ( b ) − u ( a ) v ( a ) − ∫ a b u ′ ( x ) v ( x ) d x . {\displaystyle {\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,

∫ a b u d v = u v | a b − ∫ a b v d u {\displaystyle \int _{a}^{b}udv=uv|_{a}^{b}-\int _{a}^{b}vdu} or without bounds  ∫ u d v = u v − ∫ v d u {\displaystyle \int udv=uv-\int vdu}

where  u {\displaystyle u} and  v {\displaystyle v} are functions of a variable, for instance,  x {\displaystyle x} , giving  u ( x ) {\displaystyle u(x)} and  v ( x ) {\displaystyle v(x)} .

d d x u ( x ) -> d u d x = u ′ ( x ) -> u ′ ( x ) d x = d u {\displaystyle {\frac {d}{dx}}u(x)\rightarrow {\frac {du}{dx}}=u'(x)\rightarrow u'(x)dx=du} and  ∫ d v = ∫ v ( x ) = v {\displaystyle \int dv=\int v(x)=v}

Note:  d v {\displaystyle dv} is whatever terms are not included as  u {\displaystyle u} .

### ILATE Rule

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

I – inverse trigonometric functions:  arctan ⁡ ( x ) , arcsec ⁡ ( x ) , {\displaystyle \arctan(x),\ \operatorname {arcsec}(x),} etc.
L – logarithmic functions:  ln ⁡ ( x ) , log b ⁡ ( x ) , {\displaystyle \ln(x),\ \log _{b}(x),} etc.
A – polynomials:  x 2 , 3 x 50 , {\displaystyle x^{2},\ 3x^{50},} etc.
T – trigonometric functions:  sin ⁡ ( x ) , tan ⁡ ( x ) , {\displaystyle \sin(x),\ \tan(x),} etc.
E – exponential functions:  e x , 19 x , {\displaystyle e^{x},\ 19^{x},} etc.

## Derivation

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

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

Integrating both sides with respect to  x {\displaystyle x} ,

∫ ( u ( x ) v ( x ) ) ′ d x = ∫ u ′ ( x ) v ( x ) d x + ∫ u ( x ) v ′ ( x ) d x , {\displaystyle \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 ) = ∫ u ′ ( x ) v ( x ) d x + ∫ u ( x ) v ′ ( x ) d x , {\displaystyle 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:

∫ u ( x ) v ′ ( x ) d x = u ( x ) v ( x ) − ∫ u ′ ( x ) v ( x ) d x , {\displaystyle \int u(x)v'(x)\,dx\ =\ u(x)v(x)-\int u'(x)v(x)\,dx,}

or in terms of the differentials  d u = u ′ ( x ) d x , d v = v ′ ( x ) d x , {\displaystyle \ du=u'(x)\,dx,\ \ dv=v'(x)\,dx,\quad }

∫ u ( x ) d v = u ( x ) v ( x ) − ∫ v ( x ) d u . {\displaystyle \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:

> ∫ a b u ( x ) v ′ ( x ) d x = u ( b ) v ( b ) − u ( a ) v ( a ) − ∫ a b u ′ ( x ) v ( x ) d x . {\displaystyle \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

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### Functions multiplied by one and itself

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

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

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

Let:

u = ln ⁡ ( x ) ⇒ d u = d x x {\displaystyle u=\ln(x)\ \Rightarrow \ du={\frac {dx}{x}}}
d v = d x ⇒ v = x {\displaystyle dv=dx\ \Rightarrow \ v=x}

then:

∫ ln ⁡ ( x ) d x = x ln ⁡ ( x ) − ∫ x x d x = x ln ⁡ ( x ) − ∫ 1 d x = x ln ⁡ ( x ) − x + C {\displaystyle {\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 = ∫ arctan ⁡ ( x ) d x . {\displaystyle I=\int \arctan(x)\ dx.}

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

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

Rewrite this as

∫ arctan ⁡ ( x ) ⋅ 1 d x . {\displaystyle \int \arctan(x)\cdot 1\ dx.}

Now let:

u = arctan ⁡ ( x ) ⇒ d u = d x 1 + x 2 {\displaystyle u=\arctan(x)\ \Rightarrow \ du={\frac {dx}{1+x^{2}}}}

d v = d x ⇒ v = x {\displaystyle dv=dx\ \Rightarrow \ v=x}

then

∫ arctan ⁡ ( x ) d x = x arctan ⁡ ( x ) − ∫ x 1 + x 2 d x = x arctan ⁡ ( x ) − ln ⁡ ( 1 + x 2 ) 2 + C {\displaystyle {\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

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In order to calculate

I = ∫ x cos ⁡ ( x ) d x , {\displaystyle I=\int x\cos(x)\ dx\ ,}

let:

u = x ⇒ d u = d x {\displaystyle u=x\ \Rightarrow \ du=dx}
d v = cos ⁡ ( x ) d x ⇒ v = ∫ cos ⁡ ( x ) d x = sin ⁡ ( x ) {\displaystyle dv=\cos(x)\ dx\ \Rightarrow \ v=\int \cos(x)\ dx=\sin(x)}

then:

∫ x cos ⁡ ( x ) d x = ∫ u d v = u ⋅ v − ∫ v d u = x sin ⁡ ( x ) − ∫ sin ⁡ ( x ) d x = x sin ⁡ ( x ) + cos ⁡ ( x ) + C , {\displaystyle {\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

∫ x n e x d x , ∫ x n sin ⁡ ( x ) d x , ∫ x n cos ⁡ ( x ) d x , {\displaystyle \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

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∫ x 3 e x 2 d x , {\displaystyle \int x^{3}e^{x^{2}}\,dx,}

one would set

u = x 2 , d v = x ⋅ e x 2 d x , {\displaystyle u=x^{2},\quad dv=x\cdot e^{x^{2}}\,dx,}

so that

d u = 2 x d x , v = e x 2 2 . {\displaystyle du=2x\,dx,\quad v={\frac {e^{x^{2}}}{2}}.}

Then

∫ x 3 e x 2 d x = ∫ ( x 2 ) ( x e x 2 ) d x = ∫ u d v = u v − ∫ v d u = x 2 e x 2 2 − ∫ x e x 2 d x . {\displaystyle \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

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

### Performing IBP twice

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I = ∫ e x cos ⁡ ( x ) d x . {\displaystyle I=\int e^{x}\cos(x)\ dx.}

Here, integration by parts is performed twice. First let

u = cos ⁡ ( x ) ⇒ d u = − sin ⁡ ( x ) d x {\displaystyle u=\cos(x)\ \Rightarrow \ du=-\sin(x)\ dx}
d v = e x d x ⇒ v = ∫ e x d x = e x {\displaystyle dv=e^{x}\ dx\ \Rightarrow \ v=\int e^{x}\ dx=e^{x}}

then:

∫ e x cos ⁡ ( x ) d x = e x cos ⁡ ( x ) + ∫ e x sin ⁡ ( x ) d x . {\displaystyle \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 ) ⇒ d u = cos ⁡ ( x ) d x {\displaystyle u=\sin(x)\ \Rightarrow \ du=\cos(x)\ dx}
d v = e x d x ⇒ v = ∫ e x d x = e x . {\displaystyle dv=e^{x}\ dx\ \Rightarrow \ v=\int e^{x}\ dx=e^{x}.}

Then:

∫ e x sin ⁡ ( x ) d x = e x sin ⁡ ( x ) − ∫ e x cos ⁡ ( x ) d x . {\displaystyle \int e^{x}\sin(x)\ dx=e^{x}\sin(x)-\int e^{x}\cos(x)\ dx.}

Putting these together,

∫ e x cos ⁡ ( x ) d x = e x cos ⁡ ( x ) + e x sin ⁡ ( x ) − ∫ e x cos ⁡ ( x ) d x . {\displaystyle \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 ∫ e x cos ⁡ ( x ) d x = e x [ sin ⁡ ( x ) + cos ⁡ ( x ) ] + C , {\displaystyle 2\int e^{x}\cos(x)\ dx=e^{x}{\bigl [}\sin(x)+\cos(x){\bigr ]}+C,}

which rearranges to

∫ e x cos ⁡ ( x ) d x = 1 2 e x [ sin ⁡ ( x ) + cos ⁡ ( x ) ] + C ′ {\displaystyle \int e^{x}\cos(x)\ dx={\frac {1}{2}}e^{x}{\bigl [}\sin(x)+\cos(x){\bigr ]}+C'}

## Problem Set

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1)  ∫ 2 x cos ⁡ ( 5 − 13 x ) d x {\displaystyle \int 2x\cos(5-13x)dx}

2)  ∫ x cos ⁡ ( x ) d x {\displaystyle \int x\cos(x)dx}

3)  ∫ x 3 sin ⁡ ( 3 x ) d x {\displaystyle \int x^{3}\sin(3x)dx}

4)  ∫ e 2 x cos ⁡ ( 3 x ) d x {\displaystyle \int e^{2x}\cos(3x)dx}

5)  ∫ e − x ( 2 x 2 − 3 x + 5 ) d x {\displaystyle \int e^{-x}(2x^{2}-3x+5)dx}

6)  ∫ 4 tan − 1 ⁡ ( 3 x ) d x {\displaystyle \int 4\tan ^{-1}({3 \over x})dx}

7)  ∫ x tan ⁡ ( x − 4 x 2 ) d x {\displaystyle \int x\tan(x-4x^{2})dx}

8)  ∫ x 5 cos − 1 ⁡ ( 3 x ) d x {\displaystyle \int x^{5}\cos ^{-1}(3x)dx}

9)  ∫ e 2 x 3 x − 2 d x {\displaystyle \int e^{2x}3x^{-2}dx}

10)  ∫ x ln ⁡ ( 3 x ) d x {\displaystyle \int x\ln(3x)dx}