1 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.

2 Examples

In order to calculate:

Let:

u = x, so that du = dx,

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

Then:

where C is an arbitrary constant of integration.

By repeatedly using integration by parts, integrals such as

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:

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:

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:

Putting these together, we get

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:

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:

Let:

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

v = x; dv = 1·dx

Then:

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:

Now let:

u = arctan(x); du = 1/(1+x²) dx

v = x; dv = 1·dx

Then:

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

3 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

By integrating both sides, we get

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

(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.

4 A practical note

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

This is specially useful when is a polynomial, where each consecutive derivate of f(x) is simpler, and eventually, it's a constant. In general, if 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 :

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 substitutionCalculus In mathematics, trigonometric substitution is the substitution of trigonometric functions for other expressions. One may use the trigonometric identities : : : to simplify certain integrals containing the radical expressions : : : respectively..

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