How do you find the integral of sin pi x on the interval 0 to 1? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Gió Apr 21, 2015 You have: int_0^1sin(pix)dx=-cos(pix)/pi ]_0^1=-1/pi(cos(pi)-(cos(0)))=2/pi 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 7575 views around the world You can reuse this answer Creative Commons License