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How do you integrate #int (cos(x)/sin(x)-1)dx#?

Answer 1

Elijah Abram

#ln|sinx|-x+C#

#int(cosx/sinx-1)dx=int(cotx-1)dx#

#=intcotxdx-intdx#

#=ln|sinx|-x+C#

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Answer 2

Paisley Johnson

To integrate (\int \left(\frac{\cos(x)}{\sin(x)} - 1\right)dx), you would first split it into two separate integrals:

[\int \frac{\cos(x)}{\sin(x)} dx - \int 1 , dx]

Then, integrate each part separately:

For (\int \frac{\cos(x)}{\sin(x)} dx): Use the substitution (u = \sin(x)) or (du = \cos(x) , dx) to simplify the integral. This leads to: [\int \frac{1}{u} du] which integrates to (\ln|\sin(x)| + C_1).
For (\int 1 , dx): This simply integrates to (x + C_2).

So, the final result is:

[\ln|\sin(x)| - x + 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.