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?
 A: Seeing as you intended to substitute $x = \sin t$, then it necessarily follows that  $dx = \cos t \,dt$
So $dt \neq dx$, which is what you implicitly assumed in your first transformation.
Instead, we have $$\int \sqrt {1-x^2} \,dx = \int \sqrt{1-\sin^2 t}\cos t\,dt = \int \cos^2t \,dt$$
Now, can you finish off, using an appropriate identity?
A: $$\int \sqrt{1-x^2} dx$$ Substitute $x=\sin t\Leftarrow\Rightarrow dx=\cos t \cdot dt$
$$\int \sqrt{1-\sin^2t} \cdot \cos t\cdot dt=\int (\cos t)^2 dt=\frac{t+\sin t\cos t}{2}=\frac{\arcsin(x)+x\sqrt{1-x^2}}{2}$$
A: 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.
$$\begin{align}
\int\sqrt{1-x^2}\,\mathrm{d}x
&=\int\frac{1-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-\int\frac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=\arcsin{x}-\int\frac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\
&=\arcsin{x}+x\sqrt{1-x^2}-\int\sqrt{1-x^2}\,\mathrm{d}x\\
\implies 2\int\sqrt{1-x^2}\,\mathrm{d}x&=\arcsin{x}+x\sqrt{1-x^2}+\color{grey}{constant}\\
\implies \int\sqrt{1-x^2}\,\mathrm{d}x&=\frac{\arcsin{x}+x\sqrt{1-x^2}}{2}+\color{grey}{constant}.\\
\end{align}$$
