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 GiĆ³ 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 7096 views around the world You can reuse this answer Creative Commons License