PARTS (BC TOPIC ONLY) - Advanced Methods of Integration - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC

Master AP Calculus AB & BC


CHAPTER 8. Advanced Methods of Integration


Integration by parts is a technique based completely on the Product Rule. However, it’s unlikely that you’ll recognize that familiar and happy rule once we’re done mangling it. This integrating method was made famous in the movie Stand and Deliver, when Jaime Escalante stands in front of the chalkboard and chants “Come on, it’s tic-tac-toe.” That small cameo role catapulted integration by parts to fame, and it eventually ended up on General Hospital playing a handsome gangster doctor. However, one day, everyone noticed it was just a math formula, and it was immediately fired. The formula is still very bitter.

We’ll begin by using the Product Rule to find the derivative of the expression uv, where both u and v are functions:

d(uv) = u ∙ v' + v ∙ u', or

d(uv) = u dv + v du

If you integrate both sides of the equation, you get

Finally, solve this for ∫ u dv and you get the formula for integration by parts:

The focus of this method is splitting your difficult integral into two parts: a u and a dv like the left side of the above equation. The u portion must be something you can differentiate, whereas the dv must be something you can integrate. After that, it’s all downhill.

Example 5: Evaluate

Solution: None of the methods we’ve discussed so far can handle this baby. We’ll use parts instead. First of all, set u = x (because you can easily find its derivative) and dv = cos v dx (because you can easily find its integral). It’s true that you could have set u = cos v and dv = x dx, but if at all possible, you should choose a u whose derivative, if you kept taking it again and again, would eventually equal 0. The derivative of cos x will jump back and forth between cos x and sin x without ever becoming 0.

Since u = x, du = dx, and if (Don’t worry about “+ C”s for now—we’ll take care of them later.) According to the parts formula,

and we know what u, du, v, and dv are, so plug them in.

NOTE. You’ll see why it’s important to pick a u that, through differentiation, eventually becomes 0 in Example 6.

Our original integral, on the left, equals (and can be rewritten as) the expression on the left, which contains a very simple integral in The final answer is

x sin x + cos x + C

If you don’t believe that this is the answer, take its derivative, and you get

x cos x + sin x — sin x = x cos x

which is the original integral.

That wasn’t so bad, was it? Sometimes, however, it’s less tidy. For example, if you are integrating by parts, you’d set u = x2 and dv = cos x dx. Therefore, du = 2xdx and v = sin x. According to the formula,

Do you see what’s troubling in this equation? You cannot integrate x sin x dx easily. In fact, guess what method you’ll have to use? Parts! You’ll have to set aside the x2 sin x portion for now and expand the integral using the parts formula again, this time setting u = x and dv = sin x dx. These sorts of things happen when the du term isn’t something pretty and frilly like 1, as it was in Example 5. But don’t give up hope— there’s a handy chart you can use to integrate by parts that feels like no work at all. The only limitation it has is that the u term must eventually differentiate to 0.

Example 6: Evaluate

Solution: To set up the chart, make a u column, a dv column, and a column labeled “+/— 1”. In the first row, list your u, your dv, and a “+1”. In the second row, list du, v, and a “—1”. In the third row, take another derivative, another integral, and change the sign again. Continue until the u column becomes 0, but take the signs column one row further than that. (You’ll see why in a second).

Now draw diagonal arrows beginning at the v2 and continuing down and to the right. Do this until you get to the 0 term. Multiply each of the terms along the arrow (for example, in the first arrow, you multiply x2 ∙ sin x ∙ “+1”) to get a term in the answer (x2 sin x). You will make three arrows in this chart, so the answer has three terms. (Multiplying anything by 0 results in 0, so there’s no need for a fourth arrow.) The final answer is

x2 sin x + 2x cos x — 2 sin x + C

TIP. You always start the signs column with a “+1” sign.

This method is preferred by students. In fact, look how easy Example 5 is if you use the chart:

x sin x + cos x + C

However, as wonderful as the chart is, it’s not so handy when the u term’s derivative does not eventually become 0. The final example is about as complicated as integration by parts gets.

Example 7: Evaluate

Solution: It’s wise to choose u = sin x and dv = exdx, since the integral of dv is very easy. If you do so, du = cos x dx and v = ex.

This is unfortunate. We have to use parts again to evaluate the new integral. Stick with it, though—it will pay off. For now, well ignore the ex sin x (even though it will be part of our eventual answer) and focus on As before, we set u = cos x and dv = exdx; so, du = —sin x dx and v = ex.

Watch carefully now. The original integral, is equal to ex sin x minus what we just found to be:

It looks hopeless. The original problem was and that same expression appears again on the right side of the equation! Here’s what you do: distribute that negative sign on the right-hand side and add to both sides of the equation.

To get your final answer, just divide by 2.


Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.


(Hint: So far, you know of no easy integral for ln x.)


1. If u = x2 and dv = sin 4x dx, the derivative of u will eventually become 0, so you can use a chart to find the integral.

2. It’s smart to put dv = sec x tan x dx, since the resulting v is sec x. Therefore, u = 10x and du = 1()dx. Use a chart or the formula; either works fine.

3. This one is tricky, like Example 7. Let u = cos x and dv = ex dx; therefore, du = —sin x dx and v = ex.

The rightmost integral must be evaluated using parts again, this time with u = sin x (du — cos x dx) and dv = ex dx (v = ex).

Therefore, the original integral becomes

Add to both sides of the equation to get

4. This baby is a prime candidate for the chart, with u = x3 and dv = ex dx.

5. If you don’t know an integral for In x, then you cannot set it equal to dv. So, u = ln x and Therefore, the dv must be dx; it’s the only thing left! If dv = dx, then v = x.