Question #33f91 Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Shwetank Mauria Mar 6, 2017 #int_0^(pi/4)secx^2tanxdx=1/2# Explanation: Perhaps you mean #int_0^(pi/4)secx^2tanxdx# To find this assume #u=tanx#, then #du=sec^2xdx# and Observe that as #tan0=0# and #tan(pi/4)=1#, the new limits are #0# and #1# and #int_0^(pi/4)secx^2tanxdx=int_0^1udu# = #[u^2/2]_0^1# = #1/2-0=1/2# 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 1057 views around the world You can reuse this answer Creative Commons License