On-line Math 21

On-line Math 21

5.4  Techniques of integration

Example 6
ó
õ 		exsin(x)dx

Solution

You apply parts twice to this one, too, but in a different way. First, set u = ex , and dv = sin(x)dx , giving the following table:

u = ex du = exdx
v = -cos(x) dv = sin(x)dx

Then,
ó
õ 		exsin(x)dx
=
ó
õ 		u dv
=
u v- ó
õ 		v du
=
ex(-cos(x))- ó
õ 		(-cos(x))exdx
=
-excos(x)+ ó
õ 		excos(x)dx,
which seems to be spinning our wheels, leading us back essentially where we started. Now we do it again, keeping the choice of u to be the exponential term, and dv the trigonometric:

u = ex du = exdx
v = sin(x) dv = cos(x)dx

Then, applying parts to the integral term left over from the first application,
ó
õ 		exsin(x)dx
=
-excos(x)+ ó
õ 		excos(x)dx
=
-excos(x)+ æ
è 		exsin(x)- ó
õ 		sin(x)exdx ö
ø
=
-excos(x)+exsin(x)- ó
õ 		exsin(x)dx.
You'll notice that we have now come full circle. But there is a twist. We did not come back to exactly the same term; there is a (vital) negative sign there, and also there are those other terms. Look from the first line in that long equation, to the last:
ó
õ 		exsin(x)dx = -excos(x)+exsin(x)- ó
õ 		exsin(x)dx.
You can solve that equation for òexsin(x)dx , which is, after all, what we are trying to do. Mysteriously, when you get all integrals on one side, there should be an undetermined constant on the other, reflecting the fact that the integral involves an unknown constant term:
2 ó
õ 		exsin(x)dx = -excos(x)+exsin(x)+C,
or
ó
õ 		exsin(x)dx = 1
2
 ex( sin(x)-cos(x)) +C.
There is no need to divide the +C by 2, since after all it is an undetermined constant. It is just as undetermined after you divide it by two.

Copyright (c) 2000 by David L. Johnson.

File translated from TEX by TTH, version 2.61.
On 2 Jan 2001, 14:47.