How do you integrate #int pisinpix dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Matt B. Dec 6, 2016 Using #u#-substitution, we will find a value in the function whose derivative will appear in the integrand. Thus, we will let: #u=pix# #du=pidx# So, if we plug it back in: #=intsin(u)du# #=-cos(u)+c# #=-cos(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 2948 views around the world You can reuse this answer Creative Commons License