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Establish the integral formula for (arcsin x)2

by RoRi

Establish that the following integral formula is correct:

    \[ \int (\arcsin x)^2 \, dx = x (\arcsin x)^2 - 2x + 2 \sqrt{1-x^2} \arcsin x + C. \]

Proof. We’re going to integrate by parts twice to evaluate this integral. First,

    \begin{align*} u &= (\arcsin x)^2 & du &= (2 \arcsin x) \cdot \frac{1}{\sqrt{1-x^2}} \, dx \\ dv &= dx & v &= x. \end{align*}

So, we have

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

Now, we evaluate the integral on the right using integration by parts again. Let,

    \begin{align*} u &= \arcsin x & du &= \frac{1}{\sqrt{1-x^2}} \, dx \\ dv &= \frac{x}{\sqrt{1-x^2}} \, dx & v &= -\sqrt{1-x^2}. \end{align*}

Therefore,

    \begin{align*} \int \frac{x \arcsin x}{\sqrt{1-x^2}} \, dx &= -\sqrt{1-x^2} \arcsin x + \int \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} \, dx \\ &= -\sqrt{1-x^2} \arcsin x + x + C. \end{align*}

So, putting this back into the equation above we have

    \begin{align*} \int (\arcsin x)^2 \, dx &= x(\arcsin x)^2 - 2\left(-\sqrt{1-x^2} \arcsin x + x + C\right) \\ &= x (\arcsin x)^2 - 2x + 2 \sqrt{1-x^2} \arcsin x + C. \qquad \blacksquare \end{align*}

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