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

Answer 1

Aurora Alvarado

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#

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Answer 2

Axel Alvarez

To integrate ( \int \frac{{\sin x}}{{(2 + 3\cos x)^2}} ) using substitution:

Let ( u = 2 + 3\cos x ).
Find ( du = -3\sin x , dx ).
Rewrite the integral in terms of ( u ) and ( du ).
The integral becomes ( -\frac{1}{3} \int \frac{1}{u^2} , du ).
Integrate ( \frac{1}{u^2} ) with respect to ( u ).
The result is ( -\frac{1}{3} \left(-\frac{1}{u}\right) + C ).
Substitute back ( u = 2 + 3\cos x ) into the result.
The final answer is ( \frac{1}{3(2 + 3\cos x)} + C ).

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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.