What is the anti-derivative of sin(2x)?

Answer 1

(-cos(2x)/2) + C

Answer 2

The antiderivative of sin(2x) is equal to (-1/2) cos(2x) = C
Explanation below.

You're in luck because what's inside sin is linear (i.e. x is a power of 1). This means that when you take the derivative, all you have to do is multiply by a constant, and in addition, when you take the integral, you also have to multiply by a constant. The constant produced by taking the integral (or antiderivative) is to offset the chain rule.

Recall that the derivative pattern for sin(x) goes as follows:
f(x) = sin(x)
f'(x) = cos(x)
f''(x) = -sin(x)
f'''(x) = -cos(x)
f''''(x) = -(-sin(x)) = sin(x)

As you can see, any sin or cos is sandwiched between its positive and negative counterpart.

The reasoning behind the antiderivative of sin(2x):

We know that the DERIVATIVE of sin(2x) is 2cos(2x). Therefore, if we take the ANTIDERIVATIVE, it will be LIKE the negative version of that, which is -cos(2x). If we take the derivative of that, we ALMOST have the answer we want, except for the pesky 2, so we have to offset that by putting (1/2).

Therefore, the antiderivative is -cos(2x) * (1/2) , or (-1/2)cos(2x)
Whenever taking the antiderivative, we always have to add a constant C, so the final answer is

(-1/2)cos(2x) + C

Answer 3

you mean the integral

-1/2cos(2X)

f (x) = sin(x)
f '(x) = cos(x)

If the first derivative of sin (x) = cos (x)
then the integral of cos (x) = sin (x)

The first derivative of cos (x) = -sin (x)
then the integral of sin (x) = - cos (x)

Answer 4

-(1/2)cos(2x).

Answer 5

-cos(2x)