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Let's work on the integral

$$\int\sqrt{1-x^2}\,\mathrm{d}x$$

for which an implicit change of variable will be very useful:

$$\sin t =x$$
$$\cos t \,\mathrm{d}t = \mathrm{d}x$$

Since the right hand side is just $\mathrm{d}x$, this change may always be applied. All implicit changes of variable of the type $h(t)=x$ are directly appliable; whether these changes simplify or complicate the integrand is more subtle. In this case, the change works like a charm

$$\int\sqrt{1-x^2}\,\mathrm{d}x \,\overset{\substack{\sin t =x\\ \cos t \,\mathrm{d}t = \mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \int\sqrt{1-\sin^2(t)}\cos t \,\mathrm{d}t=\int\cos^2(t) \,\mathrm{d}t $$

Our new integrand is indeed simpler: $\cos(2t)=\cos^2(t) – \sin^2(t) = 2\cos^2(t) – 1\Longrightarrow \cos^2(t)=\dfrac{1+\cos(2t)}{2}$, and

$$\int\cos^2(t) \,\mathrm{d}t=\int\dfrac{1+\cos(2t)}{2}\,\mathrm{d}t=\dfrac{1}{2}t+\dfrac{1}{4}\sin(2t)+k$$

Now we undo our change (keep in mind that $\sin(2t)=2\sin t\cos t$)

$$\dfrac{1}{2}t+\dfrac{1}{4}\sin(2t)+k \,\overset{\substack{\sin t =x\\ t=\sin^{-1} x \\ \phantom{\downarrow}}}{=}\, \dfrac{1}{2}\sin^{-1} x+\dfrac{1}{2}x\sqrt{1-x^2}+k=\dfrac{1}{2}\left(x\sqrt{1-x^2}+\sin^{-1} x\right)+k $$

Great!! Let's differentiate to check that everything is peachy

$$\left(\dfrac{1}{2}\left(x\sqrt{1-x^2}+\sin^{-1} x\right)+k\right)'=\dfrac{1}{2}\left(\left[\sqrt{1-x^2}+x\dfrac{-2x}{2\sqrt{1-x^2}}\right]+\dfrac{1}{\sqrt{1-x^2}}\right)=\dfrac{(1-x^2)-x^2+1}{2\sqrt{1-x^2}}=\sqrt{1-x^2}$$