Integration by Parts - Bruce E. Shapiro

Math 150A TOPIC 6. INTEGRATION BY PARTS
Then
∫
uv− ∫ vdu
{(
∫}} ){
e x sin x dx = −e x cos x + e x sin x − e x sin x dx
(6.19)
∫
The term with e x sin x dx appears on both sides of the equation, but
with different signs. We can add this to both sides of the equation and
divide by two to get the following:
∫
e x sin x dx = 1 2 ex (sin x − cos x) (6.20)
A common trick is to combine the method of “u-substitution” with the
method of integration by parts, as in the following example.
∫
Example 6.6 cos √ x dx We first make the substitution z = √ x, so that
dz =
dx
2 √ x . Then since dx = 2√ xdz = 2zdz, we have
∫
cos √ ∫
x dx = 2 z cos z dz
Example 6.7
(6.21a)
= 2 (cos z + z sin z) + C by equation 6.9 (6.21b)
= 2 cos √ x + 2 √ x sin √ x + C (6.21c)
∫ √ π
0
∫ √ π
0
x 3 sin(x 2 )dx First substitute z = x 2 , so that
x 3 sin(x 2 )dx =
= 1 2
∫ π
0
∫ π
0
z 3/2 sin z
z sin z dz
dz
2 √ z
The last integral is exactly like the integral in example 6.1, so that
(6.22a)
(6.22b)
∫ √ π
x 3 sin(x 2 )dx = 1 ∣ ∣∣∣
π
2 (−z cos z + sin z) = π 2
0
0
(6.22c)
Page 18 « 2012. Last revised: February 26, 2013