How do you integrate #int sinx/(2+3cosx)^2# using substitution?

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Answer 1 · 2018-03-14T11:58:33

The answer is #=1/3*(1/(2+3cosx))+C#

Let #u=2+3cosx#

#du=-3sinxdx#

#sinxdx=-1/3du#

Therefore,

#int(sinxdx)/(2+3cosx)^2=-1/3int(du)/(u^2)#

#=-1/3intu^-2du=1/3*1/u#

#=1/3*(1/(2+3cosx))+C#