How do you find the indefinite integral of #int (2sintheta+3costheta)#?

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Answer 1 · 2017-03-05T08:11:09

The answer is #=3sintheta-2costheta+C#

We need

#intsinxdx=-cosx#

#intcosxdx=sinx#

Therefore,

#int(2sin theta+3costheta) d theta#

#=2intsinthetad theta+3intcosthetad theta#

#=-2costheta+3sintheta+C#