How do you find the indefinite integral of #int (1+secpix)^2secpixtanpixdx#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Shwetank Mauria Mar 8, 2018 #int(1+secpix)^2secpixtanpixdx=(1+secpix)^3/(3pi)# Explanation: Let #u=1+secpix# then #du=pisec(pix)tan(pix)dx# and hence #int(1+secpix)^2secpixtanpixdx# = #intu^2/pidu# = #u^3/(3pi)# = #(1+secpix)^3/(3pi)# 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 1458 views around the world You can reuse this answer Creative Commons License