# How do you evaluate the integral #int dx/(1-cos3x)#?

Use now the trigonometric identity:

So that:

And as:

we get:

Substitute now:

and we have:

and undoing the substitution:

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To evaluate the integral ∫ dx/(1-cos(3x)), you can use the trigonometric identity:

1 - cos(3x) = 2sin^2(3x/2)

Then, the integral becomes:

∫ dx / (1 - cos(3x)) = ∫ dx / (2sin^2(3x/2))

Applying the trigonometric identity:

2sin^2(θ) = 1 - cos(2θ)

We get:

∫ dx / (1 - cos(3x)) = ∫ dx / (1 - cos(2 * (3x/2)))

Which simplifies to:

∫ dx / (1 - cos(3x)) = ∫ dx / (1 - cos(3x))

Now, let t = 3x/2:

dt = (3/2) dx

dx = (2/3) dt

The integral becomes:

(2/3) ∫ dt / (1 - cos(t))

Using a standard integral formula, we have:

∫ dt / (1 - cos(t)) = -2tan(t/2) + C

Substituting back t = 3x/2:

-2tan(3x/4) + C

So, the final result is:

-2tan(3x/4) + 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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