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2.3 Partielle Integration

Aus Online Mathematik Brückenkurs 2

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Contents:

  • Integration by parts.

Learning outcomes:

After this section, you will have learned to:

  • Understand the derivation of the formula for integration by parts.
  • Solve problems about integration that require integration by parts, followed by a substitution (or vice versa).

Integration by parts

To integrate products, one sometimes can make use of a method known as integration by parts. The method is based on the reverse use of the rules for differentiation of products. If f and g are two differentiable functions then the rule for products gives

D(fg)=fg+fg.

Now if one integrates both sides one gets

fg=(fg+fg)dx=fgdx+fgdx 

or after re-ordering

fgdx=fg−fgdx. 

This gives us the formula for integration by parts.

Integration by parts:

f(x)g(x)dx=F(x)g(x)−F(x)g(x)dx. 

This means in practice that one integrates a product of functions by calling one factor f and the other g, and then replaces the integral fgdx  , hopefully, by an easier integral Fgdx,   where F is a primitive function of f and g is the derivative of g.


It is important to note that the method does not always lead to an integral which is easier than the original. It may also be crucial how one chooses the functions f and g, as the following example shows.

Example 1

Determine the integral xsinxdx .

If one chooses f=x and g=sinx one gets F=x22 and g=cosx, and the formula for integration by parts gives

xsinxdx=2x2sinx−2x2cosxdx. 

The new integral on the right-hand side in this case is not easier than the original integral.

If, instead, one chooses f=sinx and g=x then F=−cosx and g=1, and

xsinxdx=−xcosx−−1cosxdx=−xcosx+sinx+C. 

Example 2

Determine the integral  x2lnxdx .

Put f=x2 and g=lnx since differentiation eliminates the logarithm when we carry out an integration by parts: F=x33 and g=1x. This gives us that

x2lnxdx=3x3lnx−3x3x1dx=3x3lnx−31x2dx=3x3lnx−313x3+C=31x3(lnx−31)+C.

Example 3

Determine the integral  x2exdx .

Put f=ex and g=x2, which gives that F=ex and g=2x, and an integration by parts gives

x2exdx=x2ex−2xexdx. 

This requires further integration by parts to solve the new integral 2xexdx . We choose in this case f=ex and g=2x, which gives F=ex and g=2

2xexdx=2xex−2exdx=2xex−2ex+C. 

The original integral thus becomes

x2exdx=x2ex−2xex+2ex+C. 

Example 4

Determine the integral  excosxdx .

In a first integration by parts, we have chosen to integrate the factor ex and differentiate the factor cosx,

excosxdx=excosx−ex(−sinx)dx=excosx+exsinxdx.

The result of this is that we essentially have replaced the factor cosx by sinx in the integral. If we therefore use integration by parts once again (integrate the ex and differentiate the sinx) we get

exsinxdx=exsinx−excosxdx. 

Thus the original integral appears here again. Summarising we have:

excosxdx=excosx+exsinx−excosxdx 

and collecting the integrals to one side gives

excosxdx=21ex(cosx+sinx)+C. 

Although integration by parts in this case did not lead to an easier integral, we arrived at an equation in which the original integral could be ”solved for”. This is not entirely unusual when the integrand is a product of trigonometric functions and / or exponential functions.

Example 5

Determine the integral  01ex2xdx .

The integral can be rewritten as

01ex2xdx=012xe−xdx. 

Substitute f=e−x and g=2x, and integrate by parts

012xe−xdx=−2xe−x10+012e−xdx=−2xe−x10+−2e−x10=(−2e−1)−0+(−2e−1)−(−2)=−e2−e2+2=2−e4.

Example 6

Determine the integral  lnx dx .

We start by performing a variable substitution u=x  which gives du=dx2x=dx2u , that is, dx=2udu,

lnxdx=lnu2udu. 

Then we integrate by parts. Put f=2u and g=lnu, which gives

lnu2udu=u2lnu−u2u1du=u2lnu−udu=u2lnu−2u2+C=xlnx−x2+C=xlnx−21+C.

Note. An alternative approach is to rewrite the initial integrand as lnx=21lnx  and then integrate by parts the product 21lnx.

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