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

Answer 1

Luke Alvarez

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#

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Answer 2

Scarlett Allison

To find the indefinite integral of ( \int (2\sin(\theta) + 3\cos(\theta)) ), you can integrate each term separately. The integral of ( \sin(\theta) ) is ( -\cos(\theta) ) and the integral of ( \cos(\theta) ) is ( \sin(\theta) ). Thus, the indefinite integral of ( 2\sin(\theta) ) is ( -2\cos(\theta) ) and the indefinite integral of ( 3\cos(\theta) ) is ( 3\sin(\theta) ). Therefore, the indefinite integral of ( \int (2\sin(\theta) + 3\cos(\theta)) ) is ( -2\cos(\theta) + 3\sin(\theta) + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.