What is the integral of #int cos^3 (x)dx# from 0 to pi/2? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Andrea S. Feb 8, 2017 #int_0^(pi/2) cos^3xdx = 2/3# Explanation: Use the idendity #cos^2x+sin^2x = 1#: #int_0^(pi/2) cos^3xdx = int_0^(pi/2) cosx cos^2xdx = int_0^(pi/2) cosx (1-sin^2x) xdx# substitute: #t=sinx# #dt = cosx dx# #x in (0,pi/2) => t in (0,1)# so: #int_0^(pi/2) cos^3xdx = int_0^1 (1-t^2)dt = [t-t^3/3]_0^1 = 2/3# 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 45158 views around the world You can reuse this answer Creative Commons License