Integration by parts 

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}</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>

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

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

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

Connection to distributions

When defining distributions, integration rather then differentiation is the fundamental operation. The derivatives of distributions are then defined so as to make integration by parts work.

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