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How do you integrate #(sinx+secx)/tanx #?

1 Answer

Mar 15, 2018

Rewrite
#int((sinx+1/cosx)/(sinx/cosx))dx#

#=int(((sinxcosx+1)/cancelcosx)/(sinx/cancelcosx))dx#

#=int(((cancelsinxcosx)/cancelsinx)+1/sinx)dx#

#=int(cosxdx)+int(cscxdx)#

Now both of these are standard integrals

#=sinx-ln(csc(x)+cot(x)) +#Constant

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