Calculus 2nd Edition Rogawski

414 CHAPTER 8 TECHNIQUES OF INTEGRATION
The key step in Integration by Parts is deciding how to write the integrand as a product
uv ′ . Keep in mind that Integration by Parts expresses ∫ uv ′ dx in terms of uv and ∫ u ′ vdx.
This is useful if u ′ v is easier to integrate than uv ′ . Here are two guidelines:
• Choose u so that u ′ is “simpler” ∫ than u itself.
• Choose v ′ so that v = v ′ dx can be evaluated.
EXAMPLE 2 Good Versus Bad Choices of u and v ′
∫
Evaluate
xe x dx.
Solution Based on our guidelines, it makes sense to write xe x = uv ′ with
• u = x (since u ′ = 1 is simpler) ∫
• v ′ = e x (since we can evaluate v = e x dx = e x + C)
Integration by Parts gives us
∫
∫
xe x dx = u(x)v(x) −
∫
u ′ (x)v(x) dx = xe x −
e x dx = xe x − e x + C
Let’s see what happens if we write xe x = uv ′ with u = e x , v ′ = x. Then
∫
u ′ (x) = e x , v(x) = xdx= 1 2 x2 + C
∫
}{{}
xe x dx = 1 2 x2 e x
uv ′ } {{ }
uv
−
∫ 1
2 x2 e x dx
} {{ }
u ′ v
This is a poor choice of u and v ′ because the integral on the right is more complicated
than our original integral.
∫
EXAMPLE 3 Integrating by Parts More Than Once Evaluate
x 2 cos xdx.
In Example 3, it makes sense to take
u = x 2 because Integration by Parts
reduces the integration of x 2 cos x to the
integration of 2x sin x, which is easier.
Solution Apply Integration by Parts a first time with u = x 2 and v ′ = cos x:
∫
∫
∫
x
} 2 cos
{{
x
}
dx = x
} 2 {{
sin x
}
− 2x
} {{
sin x
}
dx = x 2 sin x − 2 x sin xdx 2
uv ′
uv
u ′ v
Now apply it again to the integral on the right, this time with u = x and v ′ = sin x:
∫
∫
x
}
sin
{{
x
}
dx = −x
} {{
cos x
}
− (− cos x) dx = −x cos x + sin x + C
} {{ }
uv ′
uv
u ′ v
Using this result in Eq. (2), we obtain
∫
∫
x 2 cos xdx= x 2 sin x − 2 x sin xdx= x 2 sin x − 2(−x cos x + sin x) + C
= x 2 sin x + 2x cos x − 2 sin x + C
Integration by Parts applies to definite integrals:
∫ b
a
u(x)v ′ (x) dx = u(x)v(x)
∣
b
a
−
∫ b
a
u ′ (x)v(x) dx