How do you evaluate the definite integral #int 1+sinx# from #[pi/4, pi/2]#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer sjc Oct 25, 2016 #pi/4+sqrt2/2# Explanation: #int_(pi/4)^(pi/2)(1+sinx)dx# #=[x-cosx]_(pi/4)^(pi/2)# #=[x-cosx]^(pi/2)-[x-cosx]_(pi/4)# #(pi/2-cos(pi/2))-(pi/4-cos(pi/4))# #=pi/2-0-pi/4+sqrt2/2# #pi/4+sqrt2/2# 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 1270 views around the world You can reuse this answer Creative Commons License