# Trigonometric Substitutions

 Return more midterms? Rough meaning of grades: 29-34 is A 23-28 is B 17-22 is C 11-16 is D Regarding the quiz--if you do every homework problem that was assigned, you'll have a severe case of deja vu on the quiz! On the exam, we do not restrict ourselves like this, but you get to have a sheet of paper.

The first homework problem is to compute

 (5.4)

Your first idea might be to do some sort of substitution, e.g., , but is nowhere to be seen and this simply doesn't work. Likewise, integration by parts gets us nowhere. However, a technique called inverse trig substitutions'' and a trig identity easily dispenses with the above integral and several similar ones! Here's the crucial table:
 Expression Inverse Substitution Relevant Trig Identity or

Inverse substitution works as follows. If we write , then

This is not the same as substitution. You can just apply inverse substitution to any integral directly--usually you get something even worse, but for the integrals in this section using a substitution can vastly improve the situation.

If is a function, then you can even use inverse substitution for a definite integral. The limits of integration are obtained as follows.

 (5.5)

To help you understand this, note that as varies from to , the function varies from to , so is being integrated over exactly the same values. Note also that (5.3.2) once again illustrates Leibniz's brilliance in designing the notation for calculus.

Let's give it a shot with (5.3.1). From the table we use the inverse substition

We get

Wow! That was like magic. This is really an amazing technique. Let's use it again to find the area of an ellipse.

Example 5.3.1   Consider an ellipse with radii and , so it goes through and . An equation for the part of an ellipse in the first quadrant is

Thus the area of the entire ellipse is

The is because the integral computes th of the area of the whole ellipse. So we need to compute

Obvious substitution with ...? nope. Integration by parts...? nope.

Let's try inverse substitution. The table above suggests using , so . We get

 (5.6) (5.7) (5.8) (5.9)

Thus the area is

Consistency Check: If the ellipse is a circle, i.e., , this is , which is a well-known formula for the area of a circle.

Remark 5.3.2   Trigonometric substitution is useful for functions that involve , , , but not all at once!. See the above table for how to do each.

One other important technique is to use completing the square.

Example 5.3.3   Compute . We complete the square:

Thus

We do a usual substitution to get rid of the . Let , so . Then

Now we have an integral that we can do; it's almost identical to the previous example, but with (and this is an indefinite integral). Let , so . Then

Of course, we must transform back into a function in , and that's a little tricky. Use that

so that

Here we use that . Also, to compute , we draw a right triangle with side lengths and , and hypotenuse .

Example 5.3.4   Compute

To compute this, we complete the square, etc.

[[Draw triangle with sides and and hypotenuse . Then

Back to the integral, we have

William Stein 2006-03-15