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Use integration by parts to establish formulas for the integral of (1-x2)1/2

by RoRi

Prove

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

using integration by parts. Then use the identity x^2 = x^2 - 1 + 1 in the solution to deduce

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

Proof. To apply integration by parts let

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

Therefore,

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

This was the first requested formula. Then, we write x^2 = x^2 - 1 + 1 to obtain

    \begin{align*} &&\int \sqrt{1-x^2} &= x \sqrt{1-x^2} + \int \frac{x^2 - 1 + 1}{\sqrt{1-x^2}} \, dx \\[9pt] &&&= x \sqrt{1-x^2} - \int \sqrt{1-x^2} \, dx + \int \frac{1}{\sqrt{1-x^2}} \, dx \\[9pt] \implies && 2 \int \sqrt{1-x^2} \, dx &= x \sqrt{1-x^2} + \int \frac{1}{\sqrt{1-x^2}} \, dx \\[9pt] \implies && \int \sqrt{1-x^2}\, dx &= \frac{1}{2} x \sqrt{1-x^2} + \frac{1}{2} \int \frac{1}{\sqrt{1-x^2}} \, dx. \qquad \blacksquare \end{align*}

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