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.

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