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


  1. Substitute $t = \sin x$

  2. 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 :)


$$\begin{align}\int\dfrac{dt}{t-\sqrt{1-t^2}} &\stackrel{\small t=\sin x}\equiv\int\frac{\cos xdx}{\sin x-\cos x}\\ &=\frac12\left(\int\frac{\cos x+\sin x}{\sin x-\cos x}dx-\int\frac{\sin x-\cos x}{\sin x-\cos x}dx\right)\\ &=\frac12(\ln(t-\sqrt{1-t^2})-\arcsin t)+C\end{align}$$

  • $\begingroup$ How do you get to the third equality? $\endgroup$ – UserX Sep 15 '14 at 16:59
  • $\begingroup$ @UserX $d(\sin x-\cos x)=(\cos x+\sin x)dx$ $\endgroup$ – RE60K Sep 15 '14 at 17:00

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.


  • 2
    $\begingroup$ Faster wrt?${}{}$ $\endgroup$ – RE60K Sep 15 '14 at 16:34
  • $\begingroup$ I was expecting this "faster wrt OP's approach" ? $\endgroup$ – RE60K Sep 15 '14 at 16:49

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}


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