How do I find the antiderivative of #sin^7(x)cos^6(x)#?

Question

7882 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-23T09:24:57

In this way, remembering:

#int[f(x)]^nf'(x)dx=[f(x)]^(n+1)/(n+1)+c#,

#intsin^7xcos^6xdx=intsinxsin^6xcos^6xdx=#

#=intsinx(sin^2x)^3cos^6xdx=intsinx(1-cos^2x)^3cos^6xdx=#

#=intsinx(1-3cos^2x+3cos^4x-cos^6x)cos^6xdx=#

#=int(sinxcos^6x-3sinxcos^8x+3sinxcos^10x-sinxcos^12x)dx=#

#=-int(-sinx)cos^6xdx+3int(-sinx)cos^8xdx-3int(-sinx)cos^10xdx+int(-sinx)cos^12xdx=#

#=-cos^7x/7+3cos^9x/9-3cos^11x/11+cos^13x/13+c=#

#=-cos^7x/7+cos^9x/3-3cos^11x/11+cos^13x/13+c#.