MISCELLANEOUS METHODS OF INTEGRATION - 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


• Miscellaneous methods of integration

• Parts (BC topic only)

• Powers of trigonometric functions (BC topic only)

• Partial fractions (BC topic only)

• Improper integrals (BC topic only)

• Technology: Drawing derivative and integral graphs with your calculator

• Summing it up

Occasionally, you’ll encounter an integration problem that smells like trouble. AB students need only complete the first section of this chapter, whereas BC students have to plod all the way through it. Happy trails.


Most of the integration on the AP test is done using the Power Rule and u-substitution. Occasionally, the test writers (while deviously twisting their thin moustaches) will throw in a tricky integral or two and cackle uproariously. Integration is unlike differentiation in a fundamental way: Using the Power, Product, Quotient, and Chain Rules, you can differentiate just about anything that comes your way. Integration requires many more methods, some of which only work in very specific circumstances. However, don’t be discouraged. Below are five things you can try if all else has failed and you simply cannot integrate the problem at hand. One of these will help you if nothing else can.

1. Use a trigonometric substitution

Although the problem looks just about as impossible as can be, you can use the Baby Theorem to rewrite cot2v and change the integral to This is substantially easier because it is now possible. In the same way, was impossible until we rewrote it as and used u-substitution. If the problem is trigonometric, you’ve got options.

2. Split up the integral

If an integral looks too complicated, rewrite it in pieces, if possible. In the case of a fraction, rewrite each term of the numerator over the denominator, as in the following example.

Example 1: Evaluate

Solution: Although the denominator certainly looks like an arcsin is in your future, the numerator makes things too complicated. However, split this into two separate fractions, and the work is half done.

Let’s do the left integral first by u-substitution. If u = 9 — x2, du = —2xdx, and

The second integral is an arcsin problem with a = 3 and u = x. Since du = dx, you have

Therefore, the final answer is

TIP. If none of these techniques works, there’s always weeping, cursing, and breaking things. Although they won’t help you solve the problem, you’ll feel a whole lot better when you’re through.

3. Long division

If the integral at hand is a fraction made up of polynomials, and the degree of the numerator is greater than or equal to the degree of the denominator, you can use long division on the problem before you begin to simplify the integral:

NOTE. Don’t forget you are subtracting the entire second integral. That’s why it is negative.

Example 2: Evaluate

Solution: Because the degree of the numerator is greater than (or equal to) the degree of the denominator, you can simplify the problem by long division first:

We are not finished by any means, but our integral can now be rewritten as In order to finish this problem, you’ll have to separate it into pieces, just as you did in Example 1. Once separated, you get

NOTE. When you add C + C, you do not get 2C. Since each C is “some number,” when you add them, you’ll get some other number, which we also call C. Handy, eh?

You’ll use the Power Rule for Integrals, u-substitution, and arctan, respectively, to solve this, and the final answer will be

4. Complete the square

This technique is useful when you have quadratic polynomials in the denominator of your integral and, typically, only a constant in the numerator. Once you complete the square, you are able to default back to an inverse trig formula.

ALERT! When using long division, remember to use place holders of 0 for terms that are not present, like 0x2 or 0x.

Example 3: Evaluate

Solution: The quadratic in the denominator and no variables in the numerator alert us to complete the square. To do so, you’ll have to factor 2 out of the terms in the denominator (since the x2must have a coefficient of 1).

When you complete the square, you’ll add (1/2 ∙ 4)2 and subtract it in the denominator to ensure that the value of the fraction does not change.

This is now a pretty easy arctan function with u = x + 2, du = dx, and a = 1. The final answer is

5. Add and subtract (or multiply and divide) the same thing

This exercise might sound fruitless. If you add and subtract the same thing, you get zero. What’s the point? In the above example, you saw how adding and subtracting 4 allowed you to complete the square. So, it’s not a complete waste of time. This method is used most often when you are trying to do a w-substitution and the problem won’t cooperate with you. If you need something in the problem that isn’t there to finish a u-substitution, why not just add it right in (as long as you remember to subtract it as well).

Example 4: Evaluate

Solution: We can’t complete the square (not a quadratic denominator), we can’t long divide (that’s just crazy), we can’t do a trig substitution, we can’t separate (we can only separate terms in the numerator—the expression is not the same as ), and u-substitution comes up short. What are we to do? Try u-substitution again, and force it.

Let’s set u = ex + 1, so du = exdx. We’ve got a problem: There is no exdx present in the numerator—only the dx is there. So, we will add and subtract ex in the numerator like so:

(Don’t forget that there was a 1 in the denominator to start with. It wasn’t 0dx up there.) If we split this integral up, something magical transpires.

YIP. You should leave off the absolute value signs around the (ex + 1) since ex has to be positive, and that value only becomes more positive when you add 1.

The first fraction cancels out, since the numerator and denominator are equal. The second fraction is integrated with a simple u-substitution of u = ex + 1. After taking these steps, you get

and the final answer will be


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


4. Each of the following integrals varies just slightly from the others. However, each requires a completely different integration method. Discuss the method you would use to begin each.


1. (This one’s pretty tough.) If you set u = x2 + 5x + 9, du = (2x + 5)dx. In order to use u-substitution, the numerator needs to be 2x + 5. First, get the 2x up there by multiplying by 2 and 1/2 at the same time:

Now, you can add and subtract 5 to get that required 2x + 5:

Split the integral up now

and the first piece can be solved by ^-substitution (now that you have arranged it). The second integral requires you to complete the square.

2. This is a completing-the-square question. Begin by factoring the negative out of the first two terms in the denominator so that the coefficient of x2 is 1:

Now, complete the square in the denominator:

Even though it looks like you are adding 9 twice, remember that the 9 in parentheses gets multiplied by that —1, so it’s really —9.

This is the arcsin form with a = √21 and u = x + 3. Since du = dx, you can rewrite the integral as

and the answer is

3. The numerator degree is larger, so use long division (or even synthetic division since the denominator is a linear binomial).

Either way, the quotient is So, the integral can be rewritten as

which you can integrate using the Power Rule and a u-substitution of u = x — 2 to get

4. (a) Because the denominator is a quadratic with a constant numerator, you will complete the square in the denominator to integrate.

(b) You will use w-substitution to integrate, with u = x2 + 3x + 10. Since du = (2x + 3)dx, you’ll have to make the numerator match it, as you did in number 1.

(c) Because the degree of the numerator is greater than that of the denominator, long division will be your first step. I have a sinking feeling that there will be other methods required before that one is done, though.