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What is #int 1+cos2x dx #?

Answer 1

Elijah Abram

#x+1/2sin 2x + C#

if the given is #int (1+ cos 2x#) #dx#

then #int (1+ cos 2x#) #dx#=#int dx# + #1/2int cos 2x # #2*dx + C#

then #x + 1/2 sin 2x + C#

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Answer 2

Luke Alvarez

[ \int (1 + \cos(2x)) , dx ]

To integrate (1 + \cos(2x)) with respect to (x), you can use the following steps:

Integrate (1) with respect to (x), which gives (x).
Integrate (\cos(2x)) with respect to (x), which gives (\frac{1}{2}\sin(2x)) after applying the chain rule.

So, the integral becomes:

[ \int (1 + \cos(2x)) , dx = \int 1 , dx + \int \cos(2x) , dx ] [ = x + \frac{1}{2} \sin(2x) + C ]

Where (C) is the constant of integration. So, the indefinite integral of (1 + \cos(2x)) with respect to (x) is (x + \frac{1}{2} \sin(2x) + C).

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Answer 3

Christopher Allers

The integral of ( \int (1 + \cos(2x)) , dx ) equals ( x + \frac{1}{2} \sin(2x) + C ), where ( C ) is the constant of integration.

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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.