How do you integrate $int sinx/(cosx)^5$ using substitution?

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Answer 1 · 2018-06-16T11:38:29

The answer is $=1/(4cos^4x)+C$

Let $u=cosx$

$=>$ , $du=-sinxdx$

The integral is

$I=int(sinxdx)/(cos^5x)=-int(du)/u^5$

$=1/(4u^4)$

$=1/(4cos^4x)+C$

How do you integrate int sinx/(cosx)^5int sinx/(cosx)^5 using substitution?