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

1 Answer

Jun 16, 2018

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

Explanation:

Let #u=cosx#

#=>#, #du=-sinxdx#

The integral is

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

#=1/(4u^4)#

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

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