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How do you find the antiderivative of #(sinx + cosx)/tanx#?

1 Answer

Jul 27, 2016

#sinx+ln|tan(x/2)|+cosx+C#

Explanation:

# I=int(sinx+cosx)/tanxdx=int(sinx+cosx)cosx/sinxdx#

#=intcosxdx+intcos^2x/sinxdx=sinx+int(1-sin^2x)/sinxdx#

#=sinx+int{1/sinx-sinx}dx#

#=sinx+int(cscx-sinx)dx#

#:. I=sinx+ln|tan(x/2)|+cosx+C#

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