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

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Answer 1 · 2015-05-23T09:24:57

In this way, remembering:

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

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

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

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

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

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

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

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

How do I find the antiderivative of sin^7(x)cos^6(x)sin7(x)cos6(x)sin^7(x)cos^6(x)?