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

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> [!TLDR]
> substitute x = a sinθ, then integrate by parts for the deduced formula
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> **First result.** Let $I=\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}}$. Put $x=a\sin	heta$, $dx=a\cos	heta\,d	heta$.
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> $$\sqrt{a^2-x^2}=\sqrt{a^2(1-\sin^2	heta)}=a\cos	heta$$
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> $$I=\int\dfrac{a\cos	heta\,d	heta}{a\cos	heta}=\int d	heta=	heta+C=\sin^{-1}\dfrac xa+C$$
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> **Deduced result.** Let $J=\displaystyle\int\sqrt{a^2-x^2}\,dx$. Integrate by parts with $u=\sqrt{a^2-x^2}$, $dv=dx$ (so $v=x$, $du=\dfrac{-x}{\sqrt{a^2-x^2}}dx$):
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> $$J=x\sqrt{a^2-x^2}-\int x\cdot\dfrac{-x}{\sqrt{a^2-x^2}}\,dx=x\sqrt{a^2-x^2}+\int\dfrac{x^2}{\sqrt{a^2-x^2}}dx$$
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> Write $x^2=-(a^2-x^2)+a^2$:
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> $$\int\dfrac{x^2}{\sqrt{a^2-x^2}}dx=-\int\sqrt{a^2-x^2}\,dx+a^2\int\dfrac{dx}{\sqrt{a^2-x^2}}=-J+a^2\sin^{-1}\dfrac xa$$
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Q.Prove that $\displaystyle\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a} + C$ and deduce that $\displaystyle\int \sqrt{a^2-x^2}\,dx = \dfrac{x\sqrt{a^2-x^2}}{2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$ .

Concept understanding — Standard Integral Forms

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