Prove that $\displaystyle\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a} + C$ and hence deduce that $\displaystyle\int \sqrt{a^2-x^2}\,dx = \dfrac{x}{2}\sqrt{a^2-x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$.

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> [!TLDR]
> First integral: substitute $x=a\sin	heta$. Second integral: integrate by parts using $u=\sqrt{a^2-x^2},\,dv=dx$, then solve for the (repeating) integral algebraically.
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> **Part A — Prove $\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a}+C$.**
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> Let $x=a\sin	heta$, so $dx = a\cos	heta\,d	heta$, with $	heta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}ight]$ so that $\cos	heta\ge0$. Then:
> $$\sqrt{a^2-x^2} = \sqrt{a^2-a^2\sin^2	heta} = a\sqrt{1-\sin^2	heta} = a\cos	heta$$
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> $$\int\dfrac{dx}{\sqrt{a^2-x^2}} = \int\dfrac{a\cos	heta\,d	heta}{a\cos	heta} = \int d	heta = 	heta+C$$
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> Since $x=a\sin	heta \implies 	heta = \sin^{-1}\dfrac{x}{a}$:
> $$\int\dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a}+C \quad\blacksquare$$
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> **Part B — Deduce $\displaystyle\int\sqrt{a^2-x^2}\,dx = \dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+C$.**
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> Let $I=\displaystyle\int\sqrt{a^2-x^2}\,dx$. Integrate by parts with $u=\sqrt{a^2-x^2}$, $dv=dx$, so $du=\dfrac{-x}{\sqrt{a^2-x^2}}dx$, $v=x$:
> $$I = x\sqrt{a^2-x^2} - \int x\cdot\left(\dfrac{-x}{\sqrt{a^2-x^2}}ight)dx = x\sqrt{a^2-x^2} + \int\dfrac{x^2}{\sqrt{a^2-x^2}}\,dx$$
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> Write $x^2 = a^2-(a^2-x^2)$: …

Q.Prove that $\displaystyle\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a} + C$ and hence deduce that $\displaystyle\int \sqrt{a^2-x^2}\,dx = \dfrac{x}{2}\sqrt{a^2-x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$ .

Concept understanding — Standard Integral Forms

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