How do you find the antiderivative of int (cos^2x-sin^2x)dx from [0,pi/6]? Calculus Techniques of Integration Integration by Substitution 1 Answer Henry W. Oct 19, 2016 int_0^(pi/6)(cos^2x-sin^2x)dx=sqrt3/4 Explanation: int_0^(pi/6)(cos^2x-sin^2x)dx =int_0^(pi/6)(cos2x)dx =[1/2sin2x]_0^(pi/6) =1/2sin2(pi/6)-1/2sin2(0) =sqrt3/4-0=sqrt3/4 Answer link Related questions What is Integration by Substitution? How is integration by substitution related to the chain rule? How do you know When to use integration by substitution? How do you use Integration by Substitution to find intx^2*sqrt(x^3+1)dx? How do you use Integration by Substitution to find intdx/(1-6x)^4dx? How do you use Integration by Substitution to find intcos^3(x)*sin(x)dx? How do you use Integration by Substitution to find intx*sin(x^2)dx? How do you use Integration by Substitution to find intdx/(5-3x)? How do you use Integration by Substitution to find intx/(x^2+1)dx? How do you use Integration by Substitution to find inte^x*cos(e^x)dx? See all questions in Integration by Substitution Impact of this question 2143 views around the world You can reuse this answer Creative Commons License