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#