Solving Integrals by Trigonometric substitution/Procedure

Template:Mergeto Trigonometric substitution is a method for solving integrals. Not all integrals require trigonometric substitution. Check to see if a simpler method may apply to your integral before proceeding.

Deciding What to Substitute

If you are learning or practicing trigonometric substitution, and are having difficulty deciding what to to substitute; it very helpful to "guess and check." As with all math, a guess and check approach is far more useful then working backwards, or following a guide.

Trigonometric substitution is most commonly used in intergrals containing radicals. Knowing which trigonometric function to use can be difficult. What we will substitute will depend on what is underneath (or contained in) the radical.

Suppose we have the integral:

$\underset{0}{\overset{1}{\int }}x\sqrt{4-3x}dx$

First, we focus on the square root, $\sqrt{4-3x}$. It is this part of the integral which prevents us from using other, simpler methods of integration. In this case we decide to use the trigonometric identitiy:

${\sin }^2\theta +{\cos }^2\theta =1$

Before substitution, it is convienent to think of the equation in the following form:

${\cos }^2\theta =1-{\sin }^2\theta$

Now, we wish to substitute 4 sin² θ for $3x$. So we write:

$4{\sin }^2\theta =3x$

or,

$\frac{4}{3}{\sin }^2\theta =x$

Substituting

We have decided to substitute,

$\frac{4}{3}{\sin }^2\theta =x$

Now we differentiate both sides of the equation so we can subsitute in for dx. The derivative gives:

$\frac{8}{3}\sin \theta \cos \theta d\theta =dx$

Subsitution is done in two steps:

1. Substitute in for dx
2. Substitute in for x('s)

$\underset{0}{\overset{1}{\int }}x\sqrt{4-3x}dx$

1. $\underset{x=0}{\overset{x=1}{\int }}x\sqrt{4-3x}\frac{8}{3}\sin \theta \cos \theta d\theta$
2. $\underset{x=0}{\overset{x=1}{\int }}\frac{4}{3}{\sin }^2\theta \sqrt{4-3(\frac{4}{3}{\sin }^2\theta )}\frac{8}{3}\sin \theta \cos \theta d\theta$

We are done substituting becuase all of the x's are gone, and only θ's remain. Important: The original problem was given in terms of x therefore bounds on the integral (0,1) are values for x. Even though we have substituted θ for x, the bounds still refer to x. Later, we will substitute back for x, then we will deal with the bounds the normal way.

Simplifying

Initially, it seems as though the problem has become impossibly complex, but it actually simplifies a lot, and becomes managable. There are two main steps to simplifying the expression:

1. Basic algebraic simplifying. Bring constants outside of the integral, and combine like terms.
2. Substituting a trigonometric identity. This will always be the same identity used previously. In this problem we used sin² θ + cos² θ = 1

So to begin, we bring constants to the front, and combine like terms:

$\frac{32}{9}\underset{x=0}{\overset{x=1}{\int }}{\sin }^3\theta \cos \theta \sqrt{4-3(\frac{4}{3}{\sin }^2\theta )}d\theta$

Next we look at the square root. This was the entire point of our trigonometric substitution; to simlify the square root. If the square root does not simplify, then we will know that we have chosen the wrong substitution (or that we made an algebric error). Notice, how the 3 is cancelled, and the 4 is factored in the example. These two steps are crucial to solving the problem!

$\frac{32}{9}\underset{x=0}{\overset{x=1}{\int }}{\sin }^3\theta \cos \theta \sqrt{4-(4{\sin }^2\theta )}d\theta$

$\frac{32}{9}\underset{x=0}{\overset{x=1}{\int }}{\sin }^3\theta \cos \theta \sqrt{4(1-{\sin }^2\theta )}d\theta$

$\frac{32}{9}\underset{x=0}{\overset{x=1}{\int }}{\sin }^3\theta \cos \theta \sqrt{4}\sqrt{(1-{\sin }^2\theta )}d\theta$

We can preform the second step of Simplifying, which is to substitute a trigonometric identity. The point of this substitution is to eliminate the square root. (When I say, "eliminate the square root," I am refering to the square root containing the θ not the 4. If a square root of a constant remains, simply bring it outside of the integral as shown.)

$\sqrt{4}\frac{32}{9}\underset{x=0}{\overset{x=1}{\int }}{\sin }^3\theta \cos \theta \sqrt{({\cos }^2\theta )}d\theta$

$\frac{64}{9}\underset{x=0}{\overset{x=1}{\int }}{\sin }^3\theta \cos \theta (\cos \theta )d\theta$

$\frac{64}{9}\underset{x=0}{\overset{x=1}{\int }}{\sin }^3\theta {\cos }^2\theta d\theta$

Solving the Integral

Recall that the original problem asked us to `integrate`. We have performed some fancy substitution but we have not yet integrated. That is what we are going to do now.

Solving a trigonometric integral is a lesson in and of itself and, for that reason, we will not go into detail. We left off with:

$\frac{64}{9}\underset{x=0}{\overset{x=1}{\int }}{\sin }^3\theta {\cos }^2\theta d\theta$

$\frac{64}{9}\underset{x=0}{\overset{x=1}{\int }}\sin \theta {\cos }^2\theta (1-{\cos }^2\theta )d\theta$

$\frac{64}{9}\underset{x=0}{\overset{x=1}{\int }}({\cos }^2\theta \sin \theta -{\cos }^4\theta \sin \theta )d\theta$

$\frac{64}{9}\underset{x=0}{\overset{x=1}{\int }}{\cos }^2\theta \sin \theta d\theta -\frac{64}{9}\underset{x=0}{\overset{x=1}{\int }}{\cos }^4\theta \sin \theta d\theta$

$\frac{64}{9}[\frac{1}{3}{\cos }^3\theta {]}_{x=0}^{x=1}-\frac{64}{9}[\frac{1}{5}{\cos }^5\theta {]}_{x=0}^{x=1}$

$\frac{64}{27}[{\cos }^3\theta {]}_{x=0}^{x=1}-\frac{64}{45}[{\cos }^5\theta {]}_{x=0}^{x=1}$

Substituting to Get Back to x

We have successfully integrated the problem, but unfortunately, our current answer is in terms of θ; which isn't acceptable. The problem was asked in terms of x and so we should give our answer back in terms of x. To do this we refer back to our original substitution:

$4{\sin }^2\theta =3x$

We will use this equation to draw a triangle. Hooray! Drawing! But before we do that, let's rewrite our equation like this:

$\sin \theta =\frac{\sqrt{3x}}{2}$

Notice how we did not solve completly for θ. I did not want to use inverse trigonometric functions, we use the fact that "sine is equal to opposite over hypotenuse" to draw a triangle.

From the drawn triangle it can be seen that,

$\cos \theta =\frac{\sqrt{16-9x^2}}{4}$

So we can substitute into our answer so far:

$\frac{64}{9}[\frac{1}{3}{\cos }^3\theta {]}_{x=0}^{x=1}-\frac{64}{9}[\frac{1}{5}{\cos }^5\theta {]}_{x=0}^{x=1}$

$\frac{64}{9}[\frac{1}{3}(\frac{\sqrt{16-9x^2}}{4}{)}^3{]}_{x=0}^{x=1}-\frac{64}{9}[\frac{1}{5}(\frac{\sqrt{16-9x^2}}{4}{)}^5{]}_{x=0}^{x=1}$

And our answer is now in terms of x. All that is left now is plugging in the bounds.

Plugging in the Bounds

This example has not ended as well as I had hoped, so I will only say: plug in your bounds just like you always do, using the Fundamental Theroem of Calculus.