Question #7d2f5 Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Mr. Raj Nov 16, 2017 #1/2x+cosx+c# Explanation: #int(1/2-sinx )dx# #=int1/2dx-intsinxdx# #color(white)(blan)# using#(intf(x)+g(x)dx=intf(x)dx+intg(x)dx)# #=1/2x-(-cosx)+c##color(white)(blan)# #(intsinxdx=-cosx+c)# #=1/2x+cosx+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 1319 views around the world You can reuse this answer Creative Commons License