# How do you find the integral of #1/Sin^2 (x) + cos(2x)#?

I will call the left one Integral 1 and the right one Integral 2

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To find the integral of ( \frac{1}{\sin^2(x)} + \cos(2x) ), you can split the integral into two parts and then integrate each part separately.

First, let's consider ( \frac{1}{\sin^2(x)} ). Rewrite it as ( \csc^2(x) ), where ( \csc(x) ) is the cosecant function.

The integral of ( \csc^2(x) ) is ( -\cot(x) ).

Next, let's consider ( \cos(2x) ). The integral of ( \cos(2x) ) is ( \frac{1}{2}\sin(2x) ).

So, the integral of ( \frac{1}{\sin^2(x)} + \cos(2x) ) is:

[ \int \left(\frac{1}{\sin^2(x)} + \cos(2x)\right) , dx = \int \csc^2(x) , dx + \int \cos(2x) , dx ]

[ = -\cot(x) + \frac{1}{2}\sin(2x) + C ]

Where ( C ) is the constant of integration. Therefore, the integral of ( \frac{1}{\sin^2(x)} + \cos(2x) ) is ( -\cot(x) + \frac{1}{2}\sin(2x) + C ).

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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.

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.

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.

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