integral calculation is wrong. Why? $$\int \sqrt{1-x^2} dx = \int \sqrt{1-\sin^2t} \cdot dt= \int \sqrt {\cos^2 t} \cdot dt= \int \cos t \cdot dt = \sin t +C = x +C$$
The answer is wrong. Why?

Here's a way to do the integral without trigonometric substitution. All you need to know is the antiderivative $$\int\frac{\mathrm{d}x}{\sqrt{1-x^2}}=\arcsin{x}+\color{grey}{constant},$$ as well as how to integrate by parts.
$$\begin{align}
\int\sqrt{1-x^2}\,\mathrm{d}x
&=\int\frac{1-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-\int\frac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=\arcsin{x}-\int\frac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=\arcsin{x}+x\sqrt{1-x^2}-\int\sqrt{1-x^2}\,\mathrm{d}x\\
\implies 2\int\sqrt{1-x^2}\,\mathrm{d}x&=\arcsin{x}+x\sqrt{1-x^2}+\color{grey}{constant}\\
\implies \int\sqrt{1-x^2}\,\mathrm{d}x&=\frac{\arcsin{x}+x\sqrt{1-x^2}}{2}+\color{grey}{constant}.\\
\end{align}$$
