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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:

derivatives of f of x times g of x equals the derivative of f of x times g of x plus f of x times the derivative of g of x

Integrating both sides gives:

This rearranges to a more useful form:

integral of f'g = fg - integral of gf'

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

Suppose we have an integral of the form integral of h of x times k of x dx. 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:

  1. f '(x) is easy to integrate.
  2. g(x) is easy to differentiate.
  3. $\displaystyle\int\! v\, du$ is easier to compute than$\displaystyle\int\! u\, dv$.

Example

To calculate integral of  xcos(x) dx

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:

begin

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

integral of  xcos(x) dx

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!

integral of  xcos(x) dx = half x squared cos x - the integral of half x squared sin x dx

Exercise

indefinite integral of

dx =

f(x) =  and g(x) = 

so f '(x) =  g '(x) = 

Working:

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 ofln 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) = 1 over x.

Using the parts rule,

integral ofln x dx = x ln x - integral ofx1 over x dx = x ln x - integral of 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 e to the power (x squared) is 2x times  e to the power (x squared), we can integrate x times  e to the power (x squared) whereas we can't integrate e to the power (x squared).

So let f '(x) = 2x times  e to the power (x squared) which integrates to f(x) = g of x = e to the power (x squared), and g(x) = ½x2, which differentiates to g '(x) = x.

Using the parts rule:

inetgral = half (x squared) times e to the power (x squared) - half e to the power (x squared) + c

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

integral of x cubed sin x dx or integral of x to the power 8 times e to the x dx

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

Example

integral of x cubed e to the x dx

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

Then using the parts rule

= (x cubed e to the x) - the integral of 3 (x squared) e to the x dx

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

Then using the parts rule again:

the integral of 3 (x squared) e to the x dx = (3 (x squared) e to the x) - integral of 6x e to the x dx

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

Using the parts rule yet again:

 integral of 6x e to the x dx = (6x e to the x) - 6 e to the x + c

Putting it all together:

integral of x cubed e to the x dx = x3ex − 3x2ex + 6xex − 6ex + 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) = ex, so that f(x) = ex, and g(x) = sin x, which differentiates to g '(x) = cos x.

Using the parts rule:

integral of e^{x} sin (x) dxe^{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) = ex, so that f(x) = ex, and g(x) = cos x, which differentiates to g '(x) = -sin x.

Using the parts rule:

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

Combining these two, results in

integral of e^{x} sin (x) dxe^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.

indefinite integral of

dx =

f(x) =  and g(x) = 

so f '(x) =  g '(x) = 

Working:

Inverse Trigonometric Functions

Example

the integral ofarctan(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 integral of 1/(x^2 + 1) = arctan(x) + c, g '(x) = 1/(x^2 + 1)

Using the parts rule,

the integral ofarctan(x) dx = x arctan(x) − integral of x/(x^2+1) dx = x arctan(x) − ½ln(x2 + 1) + c.

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

Exercise

Calculate the integral ofarcsin(x) dx.   Show answer

answer

For completeness, the integral ofarccos(x) dx = xarccos(x) − sqrt(1-x^2)+ c.

<< Integration by Trigonometric Substitution | Integration Index

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