# How do you Integrate #cosx/(sinx)^2+sinx#?

The required integral is

This can be evaluated as two separate integrals,

Differentiating with respect to t,

which has the simple solution of

Combining both solutions,

Finally, the answer is

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To integrate (\frac{\cos(x)}{\sin^2(x)} + \sin(x)), you can separate the integral into two parts:

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

For the first part, use the substitution (u = \sin(x)), which implies (du = \cos(x)dx). The integral becomes:

[ \int \frac{1}{u^2} du = \int u^{-2} du ]

Integrating (u^{-2}) gives (-u^{-1} = -\frac{1}{u}), or in terms of (x), (-\frac{1}{\sin(x)}).

For the second part, the integral of (\sin(x)) is well-known:

[ \int \sin(x) dx = -\cos(x) ]

Therefore, the integral of the original expression is:

[ -\frac{1}{\sin(x)} - \cos(x) + C ]

where (C) is the constant of integration.

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To integrate cosx/(sinx)^2 + sinx, you can use the substitution method:

- Let u = sinx. Then du/dx = cosx.
- Rewrite the integral in terms of u: ∫(1/u^2 + 1) du.
- Integrate each term separately: ∫(1/u^2) du + ∫(1) du.
- For the first integral, use the power rule: ∫(1/u^2) du = -1/u + C.
- For the second integral, it's straightforward: ∫(1) du = u + C.
- Substitute u = sinx back into the result: -1/sinx + sinx + 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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