# Trigonometric substitution, a case I'd like to share

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• mcastillo356
mcastillo356
Gold Member
TL;DR Summary
There are some steps I haven't got a clue, and some others I should have, but ain't.
Hi, PF
First I will quote it; next the doubts and my attempt:

"In mathematics, trigonometric substitution is the replacement of trigonometric functions for other expresions. In calculus, trigonometric substitution is a technique for evaluating integrals. (...)
Case I: Integrands containing ##a^2-x^2##
Let ##x=a\sin\theta##, and use the identity ##1-\sin^2\theta=\cos^2\theta##
(...)
Example 1
In the integral
##\displaystyle\int{\displaystyle\frac{dx}{\sqrt{a^2-x^2}}}##
we may use
Then,
##\displaystyle\int{\displaystyle\frac{dx}{\sqrt{a^2-x^2}}}=\displaystyle\int{\displaystyle\frac{a\cos\theta\,d\theta}{\sqrt{a^2-a^2\sin^2\theta}}}##
The above step requires that ##a>0## and ##\cos\theta>0##. We can choose a to be the principal root of ##a^2##, and impose the restriction ##-\pi/2<\theta\<\pi/2## by using the inverse sine function." (Source: Wikipedia, "Trigonometric substitution".)

Doubts:
(i)- ##dx=a\cos\theta\,d\theta##: how is it derived?; any relationship with the Chain Rule?.
(ii)- It is required ##a>0## and ##\cos\theta>0##; every nonnegative real number has a unique nonnegative square root, called the principal square root or simply the square root. Am I right?:
(iii)- ##-1\leq x\leq 1## and ##-\pi/2\leq y\leq## are the domain and the range of ##y=\arcsin{(x)}##. I've plotted it, etc; should I explore ##x=\arcsin{(y)}##? It's clear that ##y=\cos{(x)}## is positive at the first and fourth quadrants.
Well, as you can see, I've put together doubts and attempt.
Greetings!

mcastillo356 said:
(i)- ##dx=a\cos\theta\,d\theta##: how is it derived?; any relationship with the Chain Rule?.
$$x = a\sin \theta \ \Rightarrow \ \frac{dx}{d\theta} = a\cos \theta \ \Rightarrow \ dx = (a\cos \theta)\ d\theta$$
mcastillo356 said:
(ii)- It is required ##a>0## and ##\cos\theta>0##; every nonnegative real number has a unique nonnegative square root, called the principal square root or simply the square root. Am I right?:
Technically, you could use ##|a| = \sqrt a^2## in your solution. However, you might as well assume ##a > 0##, given that you have only ##a^2## in the original integral.
mcastillo356 said:
(iii)- ##-1\leq x\leq 1## and ##-\pi/2\leq y\leq## are the domain and the range of ##y=\arcsin{(x)}##. I've plotted it, etc; should I explore ##x=\arcsin{(y)}##? It's clear that ##y=\cos{(x)}## is positive at the first and fourth quadrants.
Well, as you can see, I've put together doubts and attempt.
Greetings!
View attachment 331790
In the original integral (assuming ##a > 0##), we have ##-a \le x \le a##. Hence ##-1 \le \frac x a \le 1##. This puts ##\frac x a ## within the domain of ##\arcsin##. The range of ##\arcsin## is ##[-\frac \pi 2, \frac \pi 2]## and ##\cos## is non-negative on that range.

Last edited:
mcastillo356 and e_jane
Hi, PF, PeroK
PeroK said:
x = a\sin \theta \ \Rightarrow \ \frac{dx}{d\theta} = a\cos \theta \ \Rightarrow \ dx = (a\cos \theta)\ d\theta
No chain rule. It is just the first derivative of ##x## with respect to ##\theta##, in the Leibniz notation.
Technically, you could use ##|a| = \sqrt a^2 ## in your solution. However, you might as well assume ##a > 0##, given that you have only ##a^2## in the original integral.
Personally, I prefer the second choice. I keep in mind the concept of ##a## as the principal root of ##a^2##: there must be a transition from ##a## squared to ##a>0##.
In the original integral (assuming ##a > 0##), we have ##-a \le x \le a##. Hence ##-1 \le \frac a x \le 1##. This puts ##\frac a x## within the domain of ##\arcsin##. The range of ##\arcsin## is ##[-\frac \pi 2, \frac \pi 2]## and ##\cos## is non-negative on that range.

Shouldn't it be ##\frac x a##?
PD: Post without preview.

mcastillo356 said:
Shouldn't it be ##\frac x a##?
Yes, fixed.

mcastillo356

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