Integration by parts

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Topics in Calculus

Fundamental theorem | Function | Limits of functions | Continuity | Calculus with polynomials

Differentiation

Product rule | Quotient rule | Chain rule | Implicit differentiation | Taylor's theorem

Integration

Integration by substitution | Integration by parts | Integration by trigonometric substitution | Solids of revolution | Integration by disks | Integration by cylindrical shells | Lists of integrals

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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.

Suppose f(x) and g(x) are two continuously differentiable functions. Then the integration by parts rule states that for endpoints a, b

<math>\int_a^b f(s) g'(s)\,ds = \left[ f(x) g(x) \right]_{a}^{b} - \int_a^b g(v) f'(v)\,dv<math>

where we use the common notation

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

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

<math> 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 <math>

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

<math>\int f(x) g'(x)\,dx = f(x) g(x) - \int g(x) f'(x)\,dx<math>

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:

<math>\int u\,dv = u v - \int v\,du.<math>

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, you need to find its antiderivative g and then you still have to evaluate the resulting integral of ∫g f ' dx.

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:

<math>\int f g\,dx = f \int g\,dx - \int \left ( f' \int g\,dx \right )dx<math>

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

Table of contents

1 Application 2 Examples 3 Justification of the rule 4 A practical note

Application

The rule is helpful whenever you need to integrate a function h(x) and you are able to break it up into a product of two functions, h(x) = f(x)g(x), in such a way that you know how to differentiate f, how to integrate g, and how to deal with the resulting integral of f ' times the integral of g.

Examples

In order to calculate:

<math>\int x\cos (x) \,dx<math>

Let:

u = x, so that du = dx,

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

Then:

<math>\int x\cos (x) \,dx = \int u \,dv = uv - \int v \,du<math>

<math>\int x\cos (x) \,dx = x\sin (x) - \int \sin (x) \,dx<math>

<math>\int x\cos (x) \,dx = x\sin (x) + \cos (x) + C<math>

where C is an arbitrary constant of integration.

By repeatedly using integration by parts, integrals such as

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

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:

<math>\int e^{x} \cos (x) \,dx<math>

where, strangely enough, in the end, you don't have to do the actual integration.

This example uses integration by parts twice. First let:

u = e^x; thus du = e^xdx

v = sin(x); thus dv = cos(x)dx

Then:

<math>\int e^{x} \cos (x) \,dx = e^{x} \sin (x) - \int e^{x} \sin (x) \,dx<math>

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

u = e^x; du = e^xdx

v = -cos(x); dv = sin(x)dx

Then:

<math>\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<math>

Putting these together, we get

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

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

<math>2 \int e^{x} \cos (x) \,dx = e^{x} \left( \sin (x) + \cos (x) \right)<math>

<math>\int e^{x} \cos (x) \,dx = {e^{x} \left( \sin (x) + \cos (x) \right) \over 2}<math>

The other two famous examples are when you take something which isn't a product as a product of 1 and itself, and use integration by parts. This works if you know how to differentiate the function you want to integrate, and you also know how to integrate this derivative times x.

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

<math>\int \ln (x) \cdot 1 \,dx<math>

Let:

u = ln(x); du = 1/x dx

v = x; dv = 1·dx

Then:

<math>\int \ln (x) \,dx = x \ln (x) - \int x/x \,dx = x \ln (x) - \int 1 \,dx<math>

<math>\int \ln (x) \,dx = x \ln (x) - {x} + {C}<math>

<math>\int \ln (x) \,dx = x \left( \ln (x) - 1 \right) + C<math>

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:

<math>\int 1 \cdot \arctan (x) \,dx<math>

Now let:

u = arctan(x); du = 1/(1+x²) dx

v = x; dv = 1·dx

Then:

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

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

Justification of the rule

Integration by parts follows from the product rule of differentiation: If the two continuously differentiable functions u(x) and v(x) are given, the product rule states that

<math>(uv)' = (uv' + vu')<math>

By integrating both sides, we get

<math>uv = \int (uv' + vu') \,dx<math>

The latter integral can be written as the sum of two integrals since integration is linear:

<math>uv = \int uv' \,dx + \int vu' \,dx<math>

(the fact that u and v are continuously differentiable ensures that the two individual integrals exist.) Subtracting ∫ vu′ dx from both sides yields the desired formula of integration by parts.

<math>\int uv' \,dx = uv - \int vu' \,dx<math>

A practical note

In a general way, when you have an exponential or trigonometric function in the expression, use it as <math>v'<math>:

<math>\int f(x) e^{2x}\,dx = 2 f(x) e^{2x} - 2 \int f'(x) e^{2x}\,dx\ \dots<math>

<math>\int f(x) \cos(4x)\,dx = 4 f(x) \sin(4x) - 4 \int f'(x) \sin(4x)\,dx\ \dots<math>

This is specially useful when <math>f(x)<math> is a polynomial, where each consecutive derivate of f(x) is simpler, and eventually, it's a constant. In general, if <math>f(x)<math> is easy to derivate, this method will almost always work. Otherwise, you might have to use the substitution rule.

By contrast, when you have a logarithm or inverse trigonometric function, use it as <math>u<math>:

<math>\int f(x) \log(3x)\,dx = F(x) \log(3x) - \int F(x) \frac{3}{3x}\,dx\ \dots<math>

<math>\int f(x) \arcsin(6x)\,dx = F(x) \arcsin(6x) - \int F(x) \frac{6}{\sqrt{1 - (6x)^2}}\,dx\ \dots<math>

The objective is to reduce the inverse trigonometric function to a fraction inside the integral. If the derivate contains a radical, you may be able to apply a trigonometric substitution.

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