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How do you find the antiderivative of #(cos(2x))^2#?

Answer 1

Evelyn Agard

#F(x) = -x/2-1/8sin4x+C#

First we note that:

#cos^2(2x) = (cos(4x)-1)/2#

So, its primitive can be computed as:

#int_0^x cos^2(2t)dt = 1/2int_0^x (cos(4t)-1)dt = -x/2 + 1/8int_0^x cos(4t)d(4t) = -x/2-1/8sin4x+C#

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Answer 2

Paisley Johnson

To find the antiderivative of ( (\cos(2x))^2 ), you can use the substitution method.

Let's denote:

[ u = \cos(2x) ]

Then, find ( du ) by taking the derivative of ( u ) with respect to ( x ):

[ du = -2\sin(2x)dx ]

Now, rewrite the integral in terms of ( u ):

[ \int (\cos(2x))^2 dx = \int u^2 \cdot \frac{du}{-2\sin(2x)} ]

Simplify and integrate with respect to ( u ):

[ -\frac{1}{2} \int u^2 \csc(2x) du ]

Now integrate ( u^2 ) with respect to ( u ) and substitute back ( u = \cos(2x) ) to get the final antiderivative.

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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.