How do you find the integral of ∫sinxcos3(x)dx?? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Gió Oct 19, 2015 I found: 12cos2(x)+c Explanation: I would try considering that: d[cos(x)]=−sin(x)dx and write: ∫−d[cos(x)]cos3(x)= and integrate as if cos(x) were a simple x; =−∫cos−3(x)d[cos(x)]==−cos−2(x)−2+c=12cos2(x)+c Answer link Related questions How do I evaluate the indefinite integral ∫sin3(x)⋅cos2(x)dx ? How do I evaluate the indefinite integral ∫sin6(x)⋅cos3(x)dx ? How do I evaluate the indefinite integral ∫cos5(x)dx ? How do I evaluate the indefinite integral ∫sin2(2t)dt ? How do I evaluate the indefinite integral ∫(1+cos(x))2dx ? How do I evaluate the indefinite integral ∫sec2(x)⋅tan(x)dx ? How do I evaluate the indefinite integral ∫cot5(x)⋅sin4(x)dx ? How do I evaluate the indefinite integral ∫tan2(x)dx ? How do I evaluate the indefinite integral ∫(tan2(x)+tan4(x))2dx ? How do I evaluate the indefinite integral ∫x⋅sin(x)⋅tan(x)dx ? See all questions in Integrals of Trigonometric Functions Impact of this question 3249 views around the world You can reuse this answer Creative Commons License