# How do you find #int -cosx/arcsinx dx#?

This integral cannot be solved, using standard mathematical functions.

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To find the integral of (-\frac{\cos(x)}{\arcsin(x)} , dx), you can use integration by parts. Let (u = \arcsin(x)) and (dv = -\cos(x) , dx). Then, (du = \frac{1}{\sqrt{1 - x^2}} , dx) and (v = -\sin(x)).

Now, apply the integration by parts formula:

[\int u , dv = uv - \int v , du]

Substitute the values:

[\int -\frac{\cos(x)}{\arcsin(x)} , dx = -\arcsin(x) \cdot (-\sin(x)) - \int (-\sin(x)) \cdot \frac{1}{\sqrt{1 - x^2}} , dx]

Simplify:

[= x \sin(x) - \int \frac{\sin(x)}{\sqrt{1 - x^2}} , dx]

Now, the remaining integral can be solved by using the substitution (x = \sin(u)), (dx = \cos(u) , du). This will result in:

[= x \sin(x) - \int du]

Integrating (du) gives (u), so:

[= x \sin(x) - u + C]

Finally, replace (u) with (\arcsin(x)):

[= x \sin(x) - \arcsin(x) + C]

Where (C) is the constant of integration.

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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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