How do you use Integration by Substitution to find #intcos^3(x)*sin(x)dx#?

1 Answer
Wataru
Sep 8, 2014

By substitution,
#int cos^3xsin x dx=-1/4cos^4x+C#

Let #u=cosx#.
By taking the derivative,
#{du}/{dx}=-sinx#
By taking the reciprocal,
#{dx}/{du}=1/{-sinx}#
By multiplying by #du#,
#dx={du}/{-sinx}#

By rewriting the integral in terms of #u#,
#intcos^3xsinx dx =int u^3 sinx cdot {du}/{-sinx}#
by cancelling out #sinx#'s,
#=-int u^3 du#
by Power Rule,
#=-u^4/4+C#
by putting $u=cosx$ back in,
#=-1/4cos^4x+C#

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