How do you integrate # sqrt3sinx(cosx)^0.5#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer A. S. Adikesavan May 4, 2016 #intsqrt3 sin x (cos x)^0.5 dx=-(2sqrt3)/3( cos x )^1.5+C#. Explanation: The integral is #-sqrt3int(cos x)^0.5d(cos x)# #=-sqrt3( cos x )^(0.5+1)/(0.5+1)+C# #= -(2sqrt3)/3( cos x )^1.5+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 1985 views around the world You can reuse this answer Creative Commons License