Integration by Parts |
Integrating Products with Integration by Parts
Let's consider the statement of the product rule for derivatives. We learn that:
(u v)' = u' v + v' u
Let's apply an integral sign to each term:
The first term is the integral of a derivative and so is just uv.
Now, let's rename things: u ' = du, v ' = dv
So now we have
Now, transpose the first integral on the right to get the integration by parts formula:
which is just the integration equivalent of the product rule for derivatives.
This means that we will label one part of the integrand u, and another part
we will call dv. We will differentiate u to get du and we will antidifferentiate dv to get v. Then we'll just substitute in the parts formula.
Rule of thumb: try not to set dv equal to something that requires the power rule to get v, we wish to bring the powers down, rather than raise them. For example if we have to integrate xe^{ x} , we will not set dv equal to x -- since v would then equal (½ x²) and we will have raised powers from first to 2nd degree. An exponential antiderivative however is just an exponential function so there's no change in the exponent.
In some cases, we have no choice since part of the integrand product cannot be labeled dv.
For instance, if we have to integrate x ln x, we cannot set dv = ln x dx since we cannot
antidifferentiate ln x. It isn't anybody's derivative.
Let's do these 2 as examples.
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integration by parts | examples | practice | solutions |
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Example 1: Use integration by parts to find
u = x | dv = e^{ x} dx | |
du = dx | v = e^{ x} | |
Example 2: Use integration by parts to find .
u = ln x | dv = x dx | |
du = dx/x | v = ½ x^{ 2} | |
= ½ x^{ 2} ln x – ¼ x^{ }2 + C |
Note 1: When we set up our table with the u and the dv, notice that when there's a "d_" on the left of the equal sign, there's one on the right like in dv = x dx, or du = dx.
Note 2: When we go from u to du, we're differentiating.
When we go from dv to v, we're antidifferentiating or integrating.
Example 3: Sometimes we apply the parts formula more than once.
Use integration by parts to find .
u = x² | dv = e^{ }2 x dx | |
du = 2x dx | v = ½ e^{ }2 x | |
u_{1} = x | dv_{1} = e^{ }2 x dx | |
du_{1} = dx | v_{1} = ½e^{ }2 x | |
Note 3: when repeating the parts formula in a question it's best to use "subscripts" -- the little 1's below the line -- as in u_{1} -- when setting up the second or third equivalences.
Note 4: the antiderivative of e^{ }2 x = ½ e^{ }2 x because the derivative of e^{ }2 x = 2 e^{ }2 x so we multiply by ½ to eliminate the 2.
Example 4: Use integration by parts to find
Since this integrand is neither an even power of sec x nor an odd power of tan x, it doesn't fit any of the categories listed in standard techniques for doing trig integrals. Therefore, we'll have to improvise.
First, we break sec^{ 3} x up into sec x and sec^{ 2} x. Since sec^{ 2} x is the derivative of tan x,
we'll set that = dv and we'll let u = sec x. Then we'll apply integration by parts.
u = sec x | dv = sec^{ 2 }x dx | |
du = sec x tan x dx | v = tan x | |
use tan² x = sec² x – 1 | ||
transpose to left side | ||
integrate , divide by 2 |
Note 5: This is known as a BOOMERANG integral because it comes back at you. We started with on the left side of the equation and then ran into it again on the right side. That's how we got 2 times it on the left side in the 2nd to last step.
The same happens with integrands that include exponential functions mixed with other trig functions, especially sin x and cos x since they are both cyclical. A common approach is to name the initial integral I (in this case I = ). Then, instead of rewriting the entire expression in each step as above, we simply write "I = "
** Watch for integrals like , because though it looks like a product that requires integration by parts, a substitution of u = x^{ 2} will turn the integral into .
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integration by parts | examples | practice | solutions |
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1/ Use integration by parts on these integrals
a) | b) | c) |
d) | e) (boomerang) | f) |
2/
a) Find the area of the region bounded by y = ln x, the line x = e and the x-axis. Make a diagram.
b) Find the volume of the solid generated when the region in part (a) is revolved about the x-axis.
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integration by parts | examples | practice | solutions |
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1/ Use integration by parts on these integrals
a) | ||
u = x | dv = e^{ – x} dx | ò |
du = dx | v = – e^{ – x} | |
b) | ||
u = x | dv = sin x dx | ò |
du = dx | v = – cos x | |
– x cos x + sin x + C |
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c) | ||
u = x | dv = sec x tan x dx | |
du = dx | v = sec x | |
use integrals table for . |
x sec x – ln | sec x + tan x | + C |
d) | ||
u = ln x | dv = x^{ 1 / 2} dx | |
du = dx / x | v = 2/3 x ^{3 / 2} | |
I = 2/3 x ^{3 / 2} ln x – 4/9 x^{ }3 / 2 + C |
e) | ||
u = e^{ }– ^{x} | dv = sin x dx | |
du = – e^{ }– ^{x}^{ } dx | v = – cos x | |
u_{1} = e^{ }– ^{ x} | dv_{1} = cos x dx | |
du_{1} = – e^{ }– ^{x} dx | v_{1} = sin x | |
transpose I , factor, divide by 2 | I = – ½ e^{ }– ^{x} [cos x + sin x ] + C |
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f) | ||
u = ( ln x )^{2} | dv = dx | |
du = (2 ln x) dx / x | v = x | |
u_{1} = ln x | dv_{1} = dx | |
du_{1} = dx / x | v_{1} = x | I = x (ln x)^{ 2} – 2 x ln x + 2x + C |
2/
a) Find the area of the region bounded by y = ln x, the line x = e and the x-axis. Make a diagram.
Area = = (e ln e – e) – (ln 1 – 1) = 1 sq. unit.
(see part 2 of question (1f) above for )
b) Find the volume of the solid generated when the region in part (a) is revolved about the x-axis.
Volume =
(see question (1f) above for )
integration by parts | examples | practice | solutions |
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