Enter a quick description of your question...trigonometric integrals

Question

What is the integral of sin^3 x cos^3 x dx

Answer 1

∫sin^3xcos^3x dx

Rewrite cos^3x=cos^2x*cosx

∫sin^3xcos^2x*cosx dx

using trignometrical formula sin^2x+cos^2x=1

∫sin^3x(1-sin^2x) cosx dx

∫(sin^3x-sin^5x) cosx dx

∫sin^3xcosx dx-∫sin^5x cosx dx

let t=sinx then

dt/dx=cosx

dt=cosx dx

Rewrite the integral by substituting t=sinx and dt=cosx dx

∫t^3 dt-∫t^5 dt

t^4/4 -t^6/6

Substitute t = sinx then

Sin^4x/4-Sin^6x/6

The integral of sin^3xcos^3x dx is Sin^4x/4-Sin^6x/6