How do you evaluate the integral #int 1/(sqrt(4-x^2)dx# from 0 to 2? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Ratnaker Mehta Aug 24, 2016 #pi/2#. Explanation: Knowing that, #int1/sqrt(a^2-x^2)dx=arc sin(x/a)+C#, we have, #int_0^2 1/sqrt(4-x^2)dx=[arc sin(x/2)]_0^2=arc sin1-arc sin0=pi/2#. Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 10490 views around the world You can reuse this answer Creative Commons License