# How do you find the integral of #arccos(x)x#?

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Applying Integration in Pieces:

So,

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To find the integral of ( \text{arccos}(x) \cdot x ), you can use integration by parts. Let ( u = \text{arccos}(x) ) and ( dv = x , dx ). Then, ( du = -\frac{1}{\sqrt{1-x^2}} , dx ) and ( v = \frac{1}{2}x^2 ). Applying the integration by parts formula:

[ \int \text{arccos}(x) \cdot x , dx = uv - \int v , du ]

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

Simplify and integrate the second term. Let ( t = 1 - x^2 ), then ( dt = -2x , dx ).

[ = \frac{1}{2}x^2 \text{arccos}(x) + \frac{1}{2} \int \frac{1}{\sqrt{t}} , dt ]

[ = \frac{1}{2}x^2 \text{arccos}(x) + \frac{1}{2} \int t^{-\frac{1}{2}} , dt ]

[ = \frac{1}{2}x^2 \text{arccos}(x) + \sqrt{t} + C ]

Now, revert back to the variable ( x ).

[ = \frac{1}{2}x^2 \text{arccos}(x) + \sqrt{1 - x^2} + C ]

where ( C ) is the constant of integration.

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