Question #c3e4f Calculus Techniques of Integration Integration by Substitution 1 Answer Cem Sentin Nov 5, 2017 #int (sin6x*dx)/sinx=2/5*sin5x+2/3*sin3x+2sinx+C# Explanation: I decomposed #sin6x#, #sin6x# #=sin6x-sin2x+sin2x# #=2cos4x*sin2x+sin2x# #=sin2x*(2cos4x+1)# Hence, #(sin6x*dx)/sinx# =#[sin2x*(2cos4x+1)*dx]/sinx# =#[2sinx*cosx*(2cos4x+1)*dx]/sinx# =#2cosx*(2cos4x+1)*dx# =#int 4cos4x*cosx*dx#+#int 2cosx*dx# =#int 2cos5x*dx#+#int 2cos3x*dx#+#int 2cosx*dx# =#2/5*sin5x+2/3*sin3x+2sinx+C# Answer link Related questions What is Integration by Substitution? How is integration by substitution related to the chain rule? How do you know When to use integration by substitution? How do you use Integration by Substitution to find #intx^2*sqrt(x^3+1)dx#? How do you use Integration by Substitution to find #intdx/(1-6x)^4dx#? How do you use Integration by Substitution to find #intcos^3(x)*sin(x)dx#? How do you use Integration by Substitution to find #intx*sin(x^2)dx#? How do you use Integration by Substitution to find #intdx/(5-3x)#? How do you use Integration by Substitution to find #intx/(x^2+1)dx#? How do you use Integration by Substitution to find #inte^x*cos(e^x)dx#? See all questions in Integration by Substitution Impact of this question 1286 views around the world You can reuse this answer Creative Commons License