How do you evaluate #int cos(z)/sin^7(z) dz# for [pi/2, pi/6]? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Sasha P. Sep 26, 2015 #I=-21/2# Explanation: #sinz=t => coszdz=dt# #sin(pi/2)=1, sin(pi/6)=1/2# #I=int_(pi/2)^(pi/6) cosz/sin^7z dz= int_(1)^(1/2) dt/t^7= -1/(6t^6)|_(1)^(1/2)# #I=-1/(6*1/2^6)-(-1/(6*1))=-64/6+1/6=-63/6# #I=-21/2# 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 3257 views around the world You can reuse this answer Creative Commons License