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How do you find the antiderivative of #sin(2x)/cosx#?

Answer 1

Isabella Ackerman

Remember that #sin2x=2sinxcosx#

hence

#(sin2x)/cosx=(2sinxcosx)/(cosx)=2sinx#

Hence #int(2sinx)dx=-2cosx+c#

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Answer 2

Isabella Ackerman

To find the antiderivative of sin(2x)/cos(x), you can use the substitution method. Let ( u = \cos(x) ). Then, ( du = -\sin(x)dx ). This transforms the integral into:

[ -\int \frac{2}{u} du ]

Now, integrate ( \frac{2}{u} ) with respect to ( u ) to get ( 2\ln|u| + C ), where ( C ) is the constant of integration.

Finally, substitute back ( u = \cos(x) ) to get the antiderivative:

[ \int \frac{\sin(2x)}{\cos(x)} dx = 2\ln|\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.