# How do you integrate #int arccosx# by integration by parts method?

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To integrate (\int \arccos(x) ,dx) using the integration by parts formula, we use the formula (\int u ,dv = uv - \int v ,du), where one part is differentiated ((du)) and the other is integrated ((dv)).

Let:

- (u = \arccos(x)) → (du = -\frac{1}{\sqrt{1-x^2}} dx)
- (dv = dx) → (v = x)

Now, substitute (u), (du), (dv), and (v) into the integration by parts formula:

[ \int \arccos(x) dx = x \arccos(x) - \int x \left(-\frac{1}{\sqrt{1-x^2}}\right) dx ]

This simplifies to: [ x \arccos(x) + \int \frac{x}{\sqrt{1-x^2}} dx ]

Next, solve the integral (\int \frac{x}{\sqrt{1-x^2}} dx). Let's use a substitution method for this part:

Let (s = 1-x^2), thus (ds = -2x dx). We adjust the integral accordingly:

[ \frac{1}{2} \int \frac{-ds}{\sqrt{s}} = \frac{1}{2} \int s^{-\frac{1}{2}} (-ds) ]

Solving this integral gives:

[ \frac{1}{2} \times (-2) s^{\frac{1}{2}} = -\sqrt{s} = -\sqrt{1-x^2} ]

Bringing it all together:

[ \int \arccos(x) dx = x \arccos(x) - \sqrt{1-x^2} + 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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