In this video I go over the method of trig substitution for integrals which allows us to solve for the integrals of functions that involve circular or ellipse like equations. The method involves using the substitution rule for integrals but in reverse and this time defining the old variable as a function of the new as opposed to the other way around. This is a very useful method for evaluating integrals so make sure to watch this video!

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Trigonometric Integrals: Substitution

In finding the area of a circle or an ellipse, an integral of the form below arises:

but if it were in the form:

We could use substitution:

But as it stands the first integral is more difficult.

If we, instead, apply a trig substitution in the form:

Notice the difference between the two substitutions:

(1) The new variable, u, is a function of the old variable, x:

(2) The old variable, x, is a function of the new variable, Ѳ:

In general we can make a substitution of the form x = g(t) by using the Substitution Rule in reverse.

To make our calculations simpler, we assume that g has an inverse function; that is, g is one-to-one function (see my earlier video in the description).

In this case, we can re-write the substitution rule:

by replacing u with x and x by t:

This kind of substitution is called Inverse Substitution

We can make the inverse substitution x = a·sinѲ provided that it defines a one-to-one function. This can be accomplished by restricting Ѳ to lie in the interval [-π/2, π/2].

In the following table is a list of trigonometric substitutions that are effective for the given radical expression because of the specified trig identities. In each case the restriction on Ѳ is imposed to ensure that the function that defines the substitution is one-to-one. (These are the same intervals I used in my earlier videos on inverse trig functions)