How to find integral of #sinx/cos^2x#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Shwetank Mauria Sep 29, 2017 #intsinx/(cos^2x)dx=secx# Explanation: To find #intsinx/(cos^2x)dx#, let #t=cosx#, then as #dt=-sinxdx# #intsinx/(cos^2x)dx=int(-dt)/t^2# = #-(t^(-1))/(-1)# = #1/t# = #1/cosx# = #secx# Observe that #d/(dx)secx=secxtanx=sinx/cos^2x# 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 4398 views around the world You can reuse this answer Creative Commons License