# Trigonometric Integrals

##### Jan 21, 2016 - Calculus

This week I started back to my journey in mathematics, this time tackling Calculus II. Starting out the week was the section I ended with in Calc I, integration by parts. No sweat! However, the second section for the week was Trigonometric Integrals. I’m not certain what I was expecting exactly going into the section, however, after working a few problems I realized something very interesting: I really hadn’t learned any new skills, what I had learned was a way to leverage skills I already had to solve problems that were previously unapproachable. Let me demonstrate with finding a solution for the following integration problem:

$\int \frac{\sin \phi}{\cos^{3}\phi}\ d\phi$

### What am I doing here?

I’ve learned that the asking and answering of this question for myself is critical to solving problems. In this case, I’m trying to solve an integration problem but, truth be told, I know of know way to integrate this problem as it is. It’s not in a basic form. The only option in this circumstance is to manipulate the equation until it resembles a basic form that can be solved. So if I had to take a wild guess as to what I’m really doing here, it would be that I’m going to be manipulating this equation to get it into a basic form.

### Manipulating to integrate

To this point I’ve learned a few ways to manipulate problems:

1. Algebraic - Perhaps the simplest manipulation, manipulating the problem algebraically to simplify it into a basic form that can be solved.
2. U-Substitution - Useful for solving integrals that are composed of a function and its derivative.
3. Integration by Parts - Leveraging a proven relationship to break apart an integral into a function and a, hopefully, simpler integral to solve.

Note that at times, these manipulation strategies can be combined in order to achieve a solution. However, in the case of the problem I am trying to solve, none of these situations directly apply.

It should come as no surprise then, that all there is to solving trigonometric integrals is leveraging trigonometric manipulations to get an equation into a basic, solvable, form.

### Leveraging trigonometric identities

What am I doing now? I’m looking for a way to leverage a trig identity in order get this equation into a basic integral form:

$\int \frac{\sin \phi}{\cos^{3}\phi}\ d\phi$

In the course of the week, we learned some strategies involving sine and cosine products, but here their relationship is a quotient. A first thought might be to bring the cosine up to the denominator using negative exponents. However, one of the stipulations for these strategies is that the exponents of the trig functions must be greater than or equal to zero. So if I can’t manipulate it in this form, how to proceed? Some thought shows that while this isn’t a product of sine and cosine, with a little manipulation it can become a product of tangent and secant.

\begin{aligned} \int \frac{\sin \phi}{\cos^{3}\phi}\ d\phi& =\int \frac{\sin \phi}{\cos \phi}\frac{1}{\cos^{2}\phi}\ d\phi\\ & = \int \tan \phi\sec^{2}\phi\ d\phi \end{aligned}

Here I use the trig identities $\frac{\sin x}{\cos x} = \tan x$ and $\frac{1}{\cos x} = \sec x$ to achieve the goal of creating a product of tangent and secant functions.

In this case, because the integral is composed of products of tangent and secant functions, and the power of secant is even, I can use the trig identity $\sec^{2}x = 1 + \tan^{2}x$ in combination with u-substitution, where $u = \tan x$, to solve this problem.

\begin{aligned} \text{Let }\ u& = \tan \phi\\ du& =\sec^{2}\phi\ d\phi \end{aligned}

Now that I have values for $u$ and $du$, I can substitute them in the integral and finally continue forward with solving the problem.

\begin{aligned} & = \int \tan \phi\sec^{2}\phi\ d\phi\\ & = \int u\ du \end{aligned}

Awesome, $\int u\ du$? Totally a basic, solvable form. In this particular case, because the power on the secant was only 2, there was no need to use the identity $\sec^{2}x = 1 + \tan^{2}x$. With the integral completely in terms of $u$, the integral can be solved. Finally, $\tan \phi$ can be substituted back in for $u$.

\begin{aligned} & = \int u\ du\\ & =\frac{1}{2}u^2 + C\\ & =\frac{1}{2}\tan^{2}\phi + C \end{aligned}

There it is, the solution!

$\int \frac{\sin \phi}{\cos^{3}\phi}\ d\phi =\frac{1}{2}\tan^{2}\phi + C$

### Conclusions

This week I added another item for my checklist of potential manipulations that can be performed on integrals, in the effort to get them into a basic, solvable form. The interesting thing to me is how subtle the true skill is that is applied in most of these problems. With a section titled “Trigonometric Identities”, one might think the vital skill here is applying trig identities, and at first I would have agreed. However, after some thought, the real skill at work here that allows this to all to work is splitting apart power functions to perform different manipulations on different pieces. I will be interested to see how important this skill is going forward with the class.