Find $\int \dfrac{dt}{t-\sqrt{1-t^2}}$ Find $$\int \dfrac{dt}{t-\sqrt{1-t^2}}$$
MY APPROACH :


*

*Substitute $t = \sin x$

*Multiply numerator and denominator by $\cos x+\sin x$
then rewrite everything in terms in $\sin2x$ and $\cos2x$, we get Integrable functions.
But may be there is a better way. Can anyone think of anything smarter, thanks :)
 A: $$\begin{align}\int\dfrac{dt}{t-\sqrt{1-t^2}}
&\stackrel{\small t=\sin x}\equiv\int\frac{\cos xdx}{\sin x-\cos x}\\
&=\frac12\left(\int\frac{\cos x+\sin x}{\sin x-\cos x}dx-\int\frac{\sin x-\cos x}{\sin x-\cos x}dx\right)\\
&=\frac12(\ln(t-\sqrt{1-t^2})-\arcsin t)+C\end{align}$$
A: Faster approach; 
$$u=\sin t \cdots$$
$$\int \frac{\cos u }{\sin u- \cos u} \mathrm{d}u$$
Multiply the integrand by $$\frac{\sec^3(u)}{\sec^3(u)}$$
Then you got $$\int \frac{\sec^2(u)}{\sec^2(u) \tan(u)-\sec^2(u)} \mathrm{d}x \stackrel{\sec^2(u)=\tan^2(u)+1}{=}\int \frac{\sec^2(u)}{(\tan(u)-1) (1+\tan^2(u))} \mathrm{d}x$$
Substitute $s=\tan(u)$
Then use partial fraction decomposition.
Enjoy.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\overbrace{\color{#66f}{\large\int{\dd t \over t - \root{1 - t^{2}}}}}
^{\ds{t\ \equiv \sin\pars{x}}}\ =\
\int{\cos\pars{x}\,\dd x \over \sin\pars{x} - \cos\pars{x}}
\\[3mm]&={\root{2} \over 2}\int{\cos\pars{x}\,\dd x \over \sin\pars{x}\cos\pars{\pi/4} - \cos\pars{x}\sin\pars{\pi/4}}
={\root{2} \over 2}\
\overbrace{\int{\cos\pars{x}\,\dd x \over \sin\pars{x - \pi/4}}}
^{\ds{x - \pi/4\ \equiv\ y\ \imp\ x\ =\ y + \pi/4}}
\\[3mm]&={\root{2} \over 2}\int{\cos\pars{y + \pi/4}\,\dd y \over \sin\pars{y}}
=\half\int\bracks{\cot\pars{y} - 1}\,\dd y
=\half\bracks{\ln\pars{\sin\pars{y}} - y}
\\[3mm]&=\half\bracks{\ln\pars{\sin\pars{x} - \cos\pars{x}} - x} + \pars{~\mbox{a constant}~}
\\[3mm]&=\color{#66f}{\large\half\bracks{\ln\pars{t - \root{1 - t^{2}}} - \arcsin\pars{t}} + \pars{~\mbox{a constant}~}}
\end{align}
