# How do I find the integral #intsin^-1(x)dx# ?

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To find the integral ( \int \sin^{-1}(x) , dx ), you can use integration by parts. Let ( u = \sin^{-1}(x) ) and ( dv = dx ). Then, differentiate ( u ) to find ( du ), and integrate ( dv ) to find ( v ).

[ u = \sin^{-1}(x) ] [ du = \frac{1}{\sqrt{1 - x^2}} , dx ] [ dv = dx ] [ v = x ]

Now, apply the integration by parts formula:

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

Substitute the values:

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

The integral ( \int x \left(\frac{1}{\sqrt{1 - x^2}} , dx\right) ) can be evaluated by substitution. Let ( t = 1 - x^2 ), then ( dt = -2x , dx ).

[ -\frac{1}{2} \int \frac{1}{\sqrt{t}} , dt = -\sqrt{t} = -\sqrt{1 - x^2} ]

Finally, substitute this back into the original equation:

[ \int \sin^{-1}(x) , dx = x\sin^{-1}(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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