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Integration

Integration by Parts

This extremely useful rule is derived from the Product Rule for differentiation. The Product Rule can be written:

Integrating both sides gives:

This rearranges to a more useful form:

where f(x) and g(x) are two continuously differentiable functions.

Suppose we have an integral of the form . We can exploit the identity above by identifying one of the factors as f '(x) and the other factor as g(x).

This technique is effective when:

f '(x) is easy to integrate.

g(x) is easy to differentiate.

$\displaystyle\int\! v\, du$ is easier to compute than $\displaystyle\int\! u\, dv$ .

Example

To calculate

Let f '(x) = cos x, so integrating gives f(x) = -sin x, and g(x) = x, so differentiating gives g '(x) = 1.

Then using the parts rule

$\begin{align} \int x\cos (x) \,dx & = \int u \,dv \\ & = uv - \int v \,du \\ & = x\sin (x) - \int \sin (x) \,dx \\ & = x\sin (x) + \cos (x) + C \end{align}$

where C is a constant of integration.

We can show the process graphically:

To illustrate the care required in choosing the parts, we deliberately make a bad choice for the example above.

Let f '(x) = x, integrating gives f(x) = ½x², and g(x) = cos x, differentiating gives g '(x) = -sin x.

But when we use the rule the new integral is harder than the old one!

Exercise

The first examples we have given are very straight forward. At times it is necessary to think outside the square in order to choose f '(x) and g(x).

Example

Consider integral of ln x dx. We can use integration by parts but need to identify the factors.

ln x = 1*ln x.

We let f '(x) = 1 and g(x) = ln x then integrating gives f(x) = x and differentiating gives g '(x) = .

Using the parts rule,

integral of ln x dx = x ln x - x dx = x ln x - 1 dx + c = x ln x - x + c.

With more complex examples, there may be more factors to choose from. Some combinations will be much more difficult than others.

Example

Consider integral of x cubed e to the power (x squared) dx

This time we have more ways to split into 2 factors. Since the derivative of is , we can integrate whereas we can't integrate .

So let f '(x) = 2 which integrates to f(x) = , and g(x) = ½x², which differentiates to g '(x) = x.

Using the parts rule:

Related Harder Example

Multiple Integration by Parts - Reduction

Sometimes it is necessary to integrate by parts more than once. By repeatedly using integration by parts, integrals such as

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

Example

Let f '(x) = e^x, integration yields f(x) = e^x, and g(x) = x³, differentiation gives g '(x) = 3x².

Then using the parts rule

Again let f '(x) = e^x, so that f(x) = e^x, and g(x) = 3x², which differentiates to g '(x) = 6x.

Then using the parts rule again:

Once again let f '(x) = e^x, so that f(x) = e^x, and g(x) = 6x, which differentiates to g '(x) = 6.

Using the parts rule yet again:

Putting it all together:

= x³e^x − 3x²e^x + 6xe^x − 6e^x + c.

Periodic Solutions

An interesting example that is often seen is:

$integral of e^{x} sin (x) dx$

where the full integration does not need to be calculated. We use integration by parts twice.

Let f '(x) = e^x, so that f(x) = e^x, and g(x) = sin x, which differentiates to g '(x) = cos x.

Using the parts rule:

$integral of e^{x} sin (x) dx$ $e^{x} sin (x) - integral of e^{x} cos (x) dx$

This new integration looks very similar to the original one, but persist. To evaluate the new integral, we use integration by parts again.

Let f '(x) = e^x, so that f(x) = e^x, and g(x) = cos x, which differentiates to g '(x) = -sin x.

Using the parts rule:

$integral of e^{x} cos(x) dx$ $e^{x} cos (x) - integral of e^{x} sin (x) dx$

Combining these two, results in

$integral of e^{x} sin (x) dx$ $e^x sin (x) - e^{x} cos (x) - integral of e^{x} sin (x) dx$

The same integral appears on both sides of this equation. So we can simply solve the equation for the integral:

$2 times integral of e^{x} sin (x) dx = e^x sin (x) - e^{x} cos (x)$

$integral of e^{x} sin (x) dx = half (e^x sin (x) - e^{x} cos (x))$

It remains only to add a constant of integration.

Exercise

There are 10 types of problems found in these exercises. Do plenty of exercises until you feel confident about all types.

Inverse Trigonometric Functions

Example

arctan(x) dx, where arctan(x) is the inverse tangent function.

Let f '(x) = 1 so integrating, f (x) = x and g(x) = arctan(x). Remembering that = arctan(x) + c, g '(x) =

Using the parts rule,

arctan(x) dx = x arctan(x) − = x arctan(x) − ½ln(x² + 1) + c.

Note that we have used the method of substitution to evaluate the final integral.

Exercise

Calculate arcsin(x) dx. Show answer

answer

For completeness, arccos(x) dx = xarccos(x) − + c.

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