How do I evaluate the indefinite integral $intsin^6(x)*cos^3(x)dx$ ?

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Answer 1 · 2014-09-21T19:27:45

By substitution,

$int sin^6x cos^3x dx={sin^7x}/7-{sin^9x}/9+C$

Let us look at some details.

$int sin^6x cos^3x dx$

by pulling out $cosx$ ,

$=int sin^6x cos^2x cdot cosx dx$

by the trig identity $cos^2x=1-sin^2x$ ,

$=int sin^6x(1-sin^2x)cdot cosx dx$

by the substitution $u=sinx$ . $Rightarrow du=cosx dx$ ,

$=int u^6(1-u^2)du=int u^6-u^8 du$

by Power Rule,

$=u^7/7-u^9/9+C$

by putting $u=sinx$ back in,

$={sin^7x}/7-{sin^9x}/9+C$

How do I evaluate the indefinite integral intsin^6(x)*cos^3(x)dxintsin^6(x)*cos^3(x)dx ?