What is the integral of #int sin^2 (x).cos^2 (x) dx #? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Shwetank Mauria Jun 19, 2016 #intsin^2xcos^2xdx=x/8-(sin4x)/32+c# Explanation: As #sin2x=2sinxcosx# #intsin^2xcos^2xdx=1/4int(4sin^2xcos^2x)dx# = #1/4intsin^2(2x)dx# = #1/4int(1-cos4x)/2dx# = #x/8-1/8intcos4xdx# = #x/8-1/8xx(sin4x)/4+c# = #x/8-(sin4x)/32+c# 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 202217 views around the world You can reuse this answer Creative Commons License