How do you find the antiderivative of sin(pix) dx? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Mia May 1, 2017 " "intsin(pix)dx=-1/picos(pix)+C Explanation: Let " " u=cos(pix) " " du=-pisin(pix)dx " " -(du)/pi=sin(pix)dx " " intsin(pix)dx=int-(du)/pi=-1/piintdu=-1/pi(u)+C " " Therefore," "intsin(pix)dx=-1/picos(pix)+C Answer link Related questions How do I evaluate the indefinite integral intsin^3(x)*cos^2(x)dx ? How do I evaluate the indefinite integral intsin^6(x)*cos^3(x)dx ? How do I evaluate the indefinite integral intcos^5(x)dx ? How do I evaluate the indefinite integral intsin^2(2t)dt ? How do I evaluate the indefinite integral int(1+cos(x))^2dx ? How do I evaluate the indefinite integral intsec^2(x)*tan(x)dx ? How do I evaluate the indefinite integral intcot^5(x)*sin^4(x)dx ? How do I evaluate the indefinite integral inttan^2(x)dx ? How do I evaluate the indefinite integral int(tan^2(x)+tan^4(x))^2dx ? How do I evaluate the indefinite integral intx*sin(x)*tan(x)dx ? See all questions in Integrals of Trigonometric Functions Impact of this question 11015 views around the world You can reuse this answer Creative Commons License