Solving Integrals by Trigonometric substitution/Introduction

A Note

The prerequisites for this book are just as one would think, trigonometry, and some Calculus. As for the amount of Calculus, when most classes introduce the general method of substitution would be a good time; but really only the base knowledge of an integral is needed.

Introduction

The main purpose of this text is to explain a certain method of solving a large class of integral calculus problems, affectionally known as trigonometric substitution.

One of the reasons why trigonometry has such a bad reputation is probably because there are a lot of trigonometric identities, and they seem to have no apparent immediate application. While trigonometric substitution may or may not be considered as an answer to the latter, it certainly puts the former in good use.

The idea behind the trigonometric substitution is actually quite simple: to reduce complicated expressions involving square roots to polynomials of standard trigonometric functions. Integrals involving polynomial expressions (albeit in trigonometric functions) are much easier to solve than the ones containing square roots.

Let us demonstrate this idea in practice: consider the expression $\sqrt{1-x^2}$. Probably the most basic trigonometric identity is ${\sin }^2(\theta )+{\cos }^2(\theta )=1$ for an arbitrary angle $\theta$. If we replace x in this expression by $\sin (\theta )$, with the help of this trigonometric identity we see

$\sqrt{1-x^2}=\sqrt{1-{\sin }^2(\theta )}=\sqrt{{\cos }^2(\theta )}=\cos (\theta )$

Note that therefore $\theta =\arcsin (x)$ since we replaced $x^2$ with $si{n}^2(\theta )$ We would like to mention that technically one should write the absolute value of $\cos (\theta )$, in other words $|\cos (\theta )|$ as our final answer since $\sqrt{A^2}=|A|$ for all possible $A$. But as long as we are careful about the domain of all possible x and how $\cos (\theta )$ is used in the final computation, this simplification does not constitute a problem. However, we cannot directly interchange the simple expression $\cos (\theta )$ with the complicated $\sqrt{1-x^2}$ wherever it may appear, remembering that we did not take into consideration the fact that $dx=\cos (\theta )d\theta$ and the $d\theta$ is necessary for the integral. Thus, if we see an integral of the form

$\int \sqrt{1-x^2}dx$

we can rewrite it as

$\int \cos (\theta )d(\sin (\theta ))=\int {\cos }^2(\theta )d\theta$

by using the fact that $d(\sin (\theta )=\cos (\theta )d\theta$ and the fact that $dx=\cos (\theta )d\theta$. Since ${\cos }^2(\theta )=.5(1+\cos (2\theta ))$ our original integral reduces to: $.5\int d\theta$ + $.5\int \cos (2\theta )$

This integral = $\theta /2$ + $\sin (2\theta )$ or $\theta /2$ + $2\sin (\theta )\cos (\theta )$

We now know the integral; however, this is in terms of $\theta$ and not in terms of x, how it was given. To rewrite it in x, we simply plug in the values we already know.

$\sin (\theta )=x$ and $dx=\cos (\theta )d\theta$ and $\theta =\arcsin (x)$

By using these substitutions we find $\theta /2$ + $2\sin (\theta )\cos (\theta )$ = $\arcsin (x)/2$ + $x\sqrt{1-x^2}$