Integration of $\sqrt{1-x^2}$ by parts. Interpretation of 2nd term. I was wondering about the anti-derivative of $\sqrt{1-x^2}$
Method for solving it is given here. This solution is fine, but I was bit confused about last term in first line i.e.
$$\int \left(\frac{d}{dx}\sqrt{1-x^2}\int dx\right)dx$$
$$=\int x\cdot\frac{d}{dx}\sqrt{1-x^2}\cdot\ dx$$
Can't it be written as
$$\int x\cdot\ d\sqrt{1-x^2}$$
i.e. cancelling out both $dx$
Now if we substitute $1-x^2 = t^2$ then the integral would become
$$\int \sqrt{1-t^2}\cdot\ dt$$
Which is integral $I$ itself
This implies
$$I=\int 1\cdot\sqrt{1-x^2}dx=\int dx \sqrt{1-x^2}-\int \left(\frac{d}{dx}\sqrt{1-x^2}\int dx\right)dx$$
$$=x \sqrt{1-x^2}-I$$
Or
$$I=\frac{x}{2}\sqrt{1-x^2}+C$$
I know there is something wrong. Might be cancelling $dx$ was the wrong step. But same result is obtained by this alternate method:
$$I=\int 1\cdot\sqrt{1-x^2}dx=\int dx \sqrt{1-x^2}-\int \left(\frac{d}{dx}\sqrt{1-x^2}\int dx\right)dx$$
$$=x\sqrt{1-x^2}+\int\frac{x^2}{\sqrt{1-x^2}}dx$$
Now in second term again if we put $1-x^2=t^2$ which gives
$$dx=\frac{t}{-\sqrt{1-t^2}}dt$$
And so
$$I=x\sqrt{1-x^2}+\int\frac{1-t^2}{t}\cdot\frac{t}{-\sqrt{1-t^2}}dt$$
$$=x\sqrt{1-x^2}-\int \sqrt{1-t^2}\cdot\ dt$$
$$=x\sqrt{1-x^2}-I$$
Which again give same result
$$I=\frac{x}{2}\sqrt{1-x^2}+C$$
What step am I doing incorrectly? Is there an algebraic mistake? Please help. Thanks in advance. :)

\begin{align}
I & = \displaystyle\int \sqrt{1-x^2} \\
 & = x\sqrt{1-x^2}+\displaystyle \int \dfrac{x^2}{\sqrt{1-x^2}} \\ 
 & = x\sqrt{1-x^2}+\displaystyle \int \dfrac{x^2+1-1}{\sqrt{1-x^2}} \\
 & = x\sqrt{1-x^2}+\arcsin\left(x\right)-I \\ 
 & =\dfrac{x\sqrt{1-x^2}}{2}+\dfrac{\arcsin\left(x\right)}{2}+C
\end{align}
EDIT:
in your first method you were cancelling dx which is wrong and in your so called alternative method you assumed $\sqrt{1-t^2}=I$ which is wrong because you are working with indefinite integral here not definite 
so, 
you can't say $\displaystyle\int {f(x)} dx=\displaystyle\int {f(t)} dt$ where $t$ is some transformation of $x$
But yes in case it was definite integral then above will stand true because changing  variable i.e, from $x$ to $t$ will also change the bounds (aka limits ) in a way that value of integral(area under curve) remain the same and indefinite integrals lacks those bounds.so both integrals (integrals before change of variable and that after) wouldn't be equal 
