What is #int sin^3x/cos^6x dx#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Bio Nov 23, 2015 #int frac{sin^3x}{cos^6x} dx = -frac{1}{3cos^3x} + frac{1}{5cos^5x} + c#, where #c# is the constant of integration. Explanation: Use the substitution #u=cosx#. #frac{du}{dx} = -sinx# #int frac{sin^3x}{cos^6x} dx = int frac{cos^2x-1}{cos^6x} (-sinx) dx# #= int frac{u^2-1}{u^6} frac{du}{dx} dx# #= int (u^{-4} - u^{-6}) du# #= frac{u^{-3}}{-3} - frac{u^{-5}}{-5} + c#, where #c# is the constant of integration. #= -frac{1}{3cos^3x} + frac{1}{5cos^5x} + 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 10545 views around the world You can reuse this answer Creative Commons License