2016年4月17日 · sqrt((5+x)(5-x)) sqrt(25-x^2)=sqrt((5+x)(5-x))

2015年4月15日 · 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 …

2018年4月10日 · 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 ...

2018年1月13日 · 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?

2015年7月29日 · 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 …

2016年7月31日 · How do you integrate #int x^2/sqrt(25-x^2)# using trig substitutions? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer

2015年4月6日 · How do you find the integral of #x*sqrt(25+x^2)#? Calculus Techniques of Integration Integration by Trigonometric Substitution. 2 Answers

2016年11月1日 · How do you evaluate the limit #(x^2-25)/(x^2-4x-5)# as x approaches 5? Calculus Limits Determining Limits Algebraically. 1 Answer

2017年1月26日 · 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

2015年1月27日 · 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 ...