Apr 17, 2016 · sqrt((5+x)(5-x)) sqrt(25-x^2)=sqrt((5+x)(5-x))

Apr 15, 2015 · You must apply x=5/costheta and do the mathematics step by step. This integral does not seem solvable by other techniques, e.g. integration by part, just integration by trigonometric substitution. On integration by trigonometric substitution, you either apply trigonometry or trigonometric functions, such as the one herein. Unfortunately there is no …

Apr 10, 2018 · Solution: -5 <= x <= 5 or [-5,5] 25-x^2 >= 0 or (5+x)(5-x)>=0 Critical points are 5+x=0 or x= -5 and 5-x=0 or x= 5 f(x)=0 when x=-5 and x=5 Sign chart: When x< -5 ...

Jan 13, 2018 · When we write: f(x) = (25-x^2)^(1/2) we normally require 25-x^2 >= 0 and as a result say that the domain of f(x) is [-5, 5]. This is what I would call an implicit domain, i.e. a domain implicit in the formulation of the function rather than being explicitly specified. Why do we make the requirement that 25-x^2 >= 0?

Jul 29, 2015 · Notice how sqrt(25 + x^2) prop sqrt(a^2 + x^2), which implies x = atantheta with a = 5. If we let: x = 5tantheta dx = 5sec^2thetad theta sqrt(25 + x^2) = sqrt(5^2 + 5^2tan^2theta) = 5sectheta Then we get: = int 5tantheta*5sectheta*5sec^2thetad theta = 125int tanthetasecthetasec^2thetad theta Then, if we let: u = sectheta du = secthetatanthetad theta …

Jul 31, 2016 · How do you integrate #int x^2/sqrt(25-x^2)# using trig substitutions? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer

Apr 6, 2015 · How do you find the integral of #x*sqrt(25+x^2)#? Calculus Techniques of Integration Integration by Trigonometric Substitution. 2 Answers

Nov 1, 2016 · How do you evaluate the limit #(x^2-25)/(x^2-4x-5)# as x approaches 5? Calculus Limits Determining Limits Algebraically. 1 Answer

Jan 26, 2017 · How do you find the domain and range of #sqrt(25-x^2) #? Algebra Expressions, Equations, and Functions Domain and Range of a Function. 1 Answer

Jan 27, 2015 · Let's work on the radical and rewrite it in the following way: sqrt(a-bx^2)=sqrt(a[1-b/a x^2])=sqrt(a[1-(sqrt(b/a) x)^2])=sqrt(a)sqrt(1-(sqrt(b/a) x)^2) Now, if sin theta = sqrt(b/a) x, we can take advantage of the Pythagorean Trigonometric Identity cos^2 theta + sin^2 theta =1 <=> cos^2 theta = 1-sin^2 theta=1-(sqrt(b/a) x)^2 This means that ...