# Integration by parts

 Topics in calculus Differentiation Integration

In calculus, and more generally in mathematical analysis, integration by parts is a rule that transforms the integral of products of functions into other, possibly simpler, integrals. The rule arises from the product rule of differentiation.

## The rule

Suppose f(x) and g(x) are two continuously differentiable functions. Then the integration by parts rule states that given an interval with endpoints a, b, one has

$\int _{a}^{b}f(x)g'(x)\,dx=\left[f(x)g(x)\right]_{{a}}^{{b}}-\int _{a}^{b}f'(x)g(x)\,dx$

where we use the common notation

$\left[f(x)g(x)\right]_{{a}}^{{b}}=f(b)g(b)-f(a)g(a).$

The rule is shown to be true by using the product rule for derivatives and the fundamental theorem of calculus. Thus

 $f(b)g(b)-f(a)g(a)\,$ $=\int _{a}^{b}{\frac {d}{dx}}(f(x)g(x))\,dx$ $=\int _{a}^{b}f'(x)g(x)\,dx+\int _{a}^{b}f(x)g'(x)\,dx$

In the traditional calculus curriculum, this rule is often stated using indefinite integrals in the form

$\int f(x)g'(x)\,dx=f(x)g(x)-\int g(x)f'(x)\,dx$

or in an even shorter form, if we let u = f(x), v = g(x) and the differentials du = f′(x) dx and dv = g′(x) dx, then it is in the form in which it is most often seen:

$\int u\,dv=uv-\int v\,du.$

One can also formulate a discrete analogue for sequences, called summation by parts.

Note that the original integral contains the derivative of g; in order to be able to apply the rule, the antiderivative g must be found, and then the resulting integral ∫g f′ dx must be evaluated.

An alternative notation has the advantage that the factors of the original expression are identified as f and g, but the drawback of a nested integral:

$\int fg\,dx=f\int g\,dx-\int \left(f'\int g\,dx\right)dx$

This formula is valid whenever f is continuously differentiable and g is continuous.

## Examples

In order to calculate:

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

Let:

u = x, so that du = dx,
dv = cos(x) dx, so that v = sin(x).

Then:

 $\int x\cos(x)\,dx$ $=\int u\,dv$ $=uv-\int v\,du$
$\int x\cos(x)\,dx=x\sin(x)-\int \sin(x)\,dx$
$\int x\cos(x)\,dx=x\sin(x)+\cos(x)+C$

where C is an arbitrary constant of integration.

By repeatedly using integration by parts, integrals such as

$\int x^{{3}}\sin(x)\,dx\quad {\mbox{and}}\quad \int x^{{2}}e^{{x}}\,dx$

can be computed in the same fashion: each application of the rule lowers the power of x by one.

An interesting example that is commonly seen is:

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

where, strangely enough, in the end, the actual integration does not need to be performed.

This example uses integration by parts twice. First let:

u = ex; thus du = exdx
dv = cos(x)dx; thus v = sin(x)

Then:

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

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

u = ex; du = exdx
v = -cos(x); dv = sin(x)dx

Then:

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

Putting these together, we get

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

Notice that the same integral shows up on both sides of this equation. So we can simply add the integral to both sides to get:

$2\int e^{{x}}\cos(x)\,dx=e^{{x}}(\sin(x)+\cos(x))$
$\int e^{{x}}\cos(x)\,dx={e^{{x}}(\sin(x)+\cos(x)) \over 2}$

Two other well-known examples are when integration by parts is applied to a function expressed as a product of 1 and itself. This works if the derivative of the function is known, and the integral of this derivative times x is also known.

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

$\int \ln(x)\cdot 1\,dx$

Let:

u = ln(x); du = 1/x dx
v = x; dv = 1·dx

Then:

 $\int \ln(x)\,dx$ $=x\ln(x)-\int {\frac {x}{x}}\,dx$ $=x\ln(x)-\int 1\,dx$
$\int \ln(x)\,dx=x\ln(x)-{x}+{C}$
$\int \ln(x)\,dx=x(\ln(x)-1)+C$

where, again, C is the arbitrary constant of integration

The second example is ∫ arctan(x) dx, where arctan(x) is the inverse tangent function. Re-write this as:

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

Now let:

u = arctan(x); du = 1/(1+x2) dx
v = x; dv = 1·dx

Then:

 $\int \arctan(x)\,dx$ $=x\arctan(x)-\int {\frac {x}{1+x^{2}}}\,dx$ $=x\arctan(x)-{1 \over 2}\ln \left(1+x^{2}\right)+C$

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

## The LIATE rule

A rule of thumb for choosing which of two functions is to be u and which is to be dv is to choose u by whichever function comes first in this list:

L  the logarithmic function: ln x
I  inverse trigonometric functions: arctan x , arcsec x, etc.
A  algebraic (polynomial) functions: $x^{2},3x^{{50}}$, etc.
T  trigonometric functions: tan x, sec x, etc.
E  exponential functions: $e^{x}$, $13^{x}$, etc.

Then make dv the other function. You can remember the list by the mnemonic LIATE.

To demonstrate this rule, consider the integral

$\int x\cos x\,dx.\,$

Following the LIATE rule, u=x and dv=cos x , hence du=1 and v=sin x , which makes the integral become

$x\sin x-\int 1\sin x\,dx,\,$

which equals

$x\sin x+\cos x+C.\,$

In general, one tries to choose u and dv such that du is simpler than u and dv is easy to integrate.

## Higher dimensions

The formula for integration by parts can be extended to functions of several variables. Instead of an interval one needs to integrate over a n-dimensional set. Also, one replaces the derivative with a partial derivative.

More specifically, given two continuously differentiable functions u and v defined on a n-dimensional bounded set Ω with piecewise smooth boundary Γ, one has

$\int _{{\Omega }}{\frac {\partial u}{\partial x_{i}}}v\,d\Omega =\int _{{\Gamma }}uv\,{\mathbf {n}}_{i}\,d\Gamma -\int _{{\Omega }}u{\frac {\partial v}{\partial x_{i}}}\,d\Omega$

where n is the outer surface normal to Γ, ni is its i-th component, and i is between 1 and n. Adding these up with i from 1 to n with v replaced by vi, one obtains

$\int _{{\Omega }}\nabla u\cdot {\mathbf {v}}\,d\Omega =\int _{{\Gamma }}u\,{\mathbf {v}}\cdot {\mathbf {n}}\,d\Gamma -\int _{\Omega }u\,\nabla \cdot {\mathbf {v}}\,d\Omega$

where v is a vector-valued function with components v1, ..., vn. For u the constant function 1, one gets the divergence theorem.

Error creating thumbnail: File missing