How do you integrate #int (2sinx+3cosx)dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Alan N. Feb 27, 2017 #3sinx-2cosx+C# Explanation: #int (2sinx+3cosx)dx= 2*int sinx dx + 3* int cosx dx# Applying standard integrals: #= 2*(-cosx) +3*sinx +C# #=3sinx-2cosx+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 11717 views around the world You can reuse this answer Creative Commons License