- #1

mcastillo356

Gold Member

- 587

- 308

- 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##

(...)

In the integral

##\displaystyle\int{\displaystyle\frac{dx}{\sqrt{a^2-x^2}}}##

we may use

##x=a\sin\theta,\quad{dx=a\cos\theta\,d\theta,\quad{\theta=\arcsin{\displaystyle\frac{x}{a}}}}##

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}}}##

##\qquad{=\displaystyle\int{\displaystyle\frac{a\cos\theta\,d\theta}{\sqrt{a^2(1-\sin^2\theta)}}}}##

##\qquad{=\displaystyle\frac{a\cos\theta\,d\theta}{\displaystyle\sqrt{a^2\cos^2\theta}}}##

##\qquad{=\displaystyle\int{d\theta}}##

##\qquad{=\theta+C}##

##\qquad{=\arcsin{\displaystyle\frac{x}{a}}+C}##

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!

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

##x=a\sin\theta,\quad{dx=a\cos\theta\,d\theta,\quad{\theta=\arcsin{\displaystyle\frac{x}{a}}}}##

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}}}##

##\qquad{=\displaystyle\int{\displaystyle\frac{a\cos\theta\,d\theta}{\sqrt{a^2(1-\sin^2\theta)}}}}##

##\qquad{=\displaystyle\frac{a\cos\theta\,d\theta}{\displaystyle\sqrt{a^2\cos^2\theta}}}##

##\qquad{=\displaystyle\int{d\theta}}##

##\qquad{=\theta+C}##

##\qquad{=\arcsin{\displaystyle\frac{x}{a}}+C}##

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!