# How do you calculate the antiderivative of #sin(2x)/cos(x) dx#?

The integral/antiderivative you require is

The easiest way to solve it is to use trigonometric identities and expand the numerator, like this

This transforms the integral in this way

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To calculate the antiderivative of ( \frac{\sin(2x)}{\cos(x)} ) with respect to ( x ), you can use a u-substitution. Let ( u = \cos(x) ), then ( du = -\sin(x) dx ).

Substitute ( u ) and ( du ) into the integral: [ \int \frac{\sin(2x)}{\cos(x)} dx = \int \frac{2\sin(x)\cos(x)}{\cos(x)} dx ] [ = \int 2\sin(x) dx ]

Integrate ( 2\sin(x) ) with respect to ( x ): [ = -2\cos(x) + C ]

So, the antiderivative of ( \frac{\sin(2x)}{\cos(x)} ) is ( -2\cos(x) + C ), where ( C ) is the constant of integration.

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To calculate the antiderivative of ( \frac{\sin(2x)}{\cos(x)} ) with respect to ( x ), you can use trigonometric identities to simplify the expression. Specifically, you can use the double-angle identity for sine, which states that ( \sin(2x) = 2\sin(x)\cos(x) ).

So, ( \frac{\sin(2x)}{\cos(x)} ) becomes ( \frac{2\sin(x)\cos(x)}{\cos(x)} ), which simplifies to ( 2\sin(x) ).

The antiderivative of ( 2\sin(x) ) with respect to ( x ) is ( -2\cos(x) + C ), where ( C ) is the constant of integration.

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