Integration Tricks I
Consider the following types of integrals.
Ú xe
-2x
dx
2 3x
Úx e
4 5x
dx
Úx e
dx .
Below is a trick that will let you evaluate all of the above with one integration. Consider
the following function defined by an integral (Take the constant of integration to be
zero). The independent variable of the function is a .
I(a ) =
ax
Úe
1 ax
e .
a
dx =
As you can see this integral is very easy to evaluate. Now consider the derivative of the
above function with respect to a . This will give us the anti-derivative of another
integral.
d
I(a ) =
da
Ú da e
d
d
I(a ) =
da
Ú x e a x dx =
ax
dx =
d Ê 1 a xˆ
e ¯
da Ë a
ea x È
1
x- ˘
a Î
a˚
If you were doing this for a specific problem you would now set a to the desired value.
For example if you were doing the first integral on the top of the page you would set a
equal to -2.
Ú xe
-2x
e -2 x È
1
dx = x + ˘.
2 Î
2˚
Lets look at the second derivative of I with respect to a .
d2
I(a ) =
da 2
d2 ax
d2 Ê 1 a x ˆ
e
dx
=
e
Ú da 2
¯
da 2 Ë a
d2
I(a ) =
da 2
2 ax
Ú x e dx =
ea x È 2
2x
2
x + 2˘
a Î
a
a ˚
If you were doing this for a specific problem you would now set a to the desired value.
For example if you were doing the second integral on the top of the page you would set
a equal to 3.
2 3x
Ú x e dx =
e3 x È 2
2x
2
x + ˘
3 Î
3
9˚
To do the third integral on the top of the previous page you would look at the fourth
derivative of I.
Lets now consider integrals of the following form.
Úe
2x
3x
sin(3x) dx
Úe
cos(4x) dx
Úxe
2x
sin(3x) dx
2 2x
Úx e
sin(3x ) dx
Once again we will evaluate all integrals at the same time. Recall the Euler identity.
e iq = cos(q ) + isin(q )
Now consider the following integral,
I =
Úe
ax ibx
e dx =
Real(I) =
Úe
ax
Úe
ax
[cos(bx ) + i sin(bx)]dx = Ú e(a + ib)x dx
cos(bx)dx
Imaginary(I) =
Úe
ax
sin(bx)dx
I is relatively easy to evaluate (it is just an exponential integral),
I =
1
(
)
e a + ib x .
a + ib
To obtain the real and imaginary parts we need to rationalize the denominator.
(
)
I = e a + ib x
1
a - ib
(
) a - ib
= e a + ib x 2
a + ib a - ib
a + b2
Now we need to expand the complex exponential in terms of sin and cos.
a - ib
(cos(bx) + i sin(bx))
a2 + b 2
eax
I = 2
[(acos(bx) + bsin(bx)) + i (asin(bx) - bcos(bx))]
a + b2
I = e ax
We now have the results that we wanted.
e ax
( acos(bx) + bsin(bx))
a 2 + b2
e ax
(asin(bx) - bcos(bx))
Imaginary(I) = Ú e ax sin(bx)dx = 2
a + b2
Real(I) =
ax
Ú e cos(bx)dx =
To evaluate the first integral at the top of the previous page we set a = 2 and b = 3 and
use the sin integral. To evaluate the second integral at the top of the page we set a = 3
and b = 4 and use the cos integral.
e 3x
(3cos(4x) + 4sin(4x))
25
e2x
2x
Ú e sin(3x)dx = 13 ( 2sin(3x) - 3cos(3x))
3x
Ú e cos(4x)dx =
To evaluate the third integral on the top of the page you would need to compute the first
derivative with respect to ‘a’ of the sin integral and then set a = 2 and b = 3.
e ax
(asin(bx) - bcos(bx))
a2 + b2
d
d
e ax
ax
e
sin(bx)dx
=
( asin(bx) - bcos(bx))
da Ú
da a 2 + b 2
e ax ÈÊ
2a ˆ
˘
ax
xe
sin(bx)dx
=
2
2 ÍË x 2
2 ¯ (asin(bx) - bcos(bx) ) + sin(bx)˙
Ú
a +b Î
a +b
˚
ax
Ú e sin(bx)dx =
After setting a = 2 and b = 3 we obtain the desired result.
2x
Ú xe sin(3x)dx =
e2x È Ê
4ˆ
( 2sin(3x) - 3cos(3x) ) + sin(3x)˘˙
x
Í
Ë
¯
13 Î
13
˚
The above result would have been very time consuming without the trick.