Evaluate $\int\frac{dt}{t-\sqrt{1-t^2}}$

Question

How to evaluate the integrals $$\int\frac{dt}{t-\sqrt{1-t^2}}$$

I try to use $t=\sin x$ to do substitution, then $dt=\cos xdx$, so $$\int\frac{dt}{t-\sqrt{1-t^2}}=\int\frac{\cos xdx}{\sin x-\cos x}$$ but I cannot go on to handle the integral.

Answer 1

If you want to continue in the way you have started, let $$C=\int\frac{\cos x}{\sin x-\cos x}dx$$ and $$S=\int\frac{\sin x}{\sin x-\cos x}dx$$

Then $$S-C=\int 1 dx$$ and $$S+C=\int\frac{\sin x+\cos x}{\sin x -\cos x}dx=\ln|\sin x -\cos x|+c$$

So you can deduce an expression for $C$.

Answer 2

$$ \int \frac{dt}{t-\sqrt{1-t^2}} \\ = \int \frac{dt}{t-t \sqrt{\frac{1}{t^2}-1}} = \int \frac{dt}{t \big(1 - \sqrt{\frac{1}{t^2}-1}\big)} \\ = \int \frac{t^2}{t^3 \big(1 - \sqrt{\frac{1}{t^2}-1}\big)}dt $$ Make the substitution $\frac{1}{t^2}-1 = x^2 \implies t^2=\frac{1}{x^2+1} \implies \frac{-2}{t^3}dt=2xdx$

$$ \int \frac{\frac{1}{x^2+1} \cdot (-xdx)}{1-x} = \int \frac{x}{(x^2+1)(x-1)}dx \\ $$ Use Partial fractions decomposition to get

$$ \int \Big(\frac{1}{2(x-1)} -\frac{(x-1)}{2(x^2+1)}\Big)dx = \frac{1}{2} \int\frac{1}{x-1} - \frac{\frac{1}{2}(2x)-1}{x^2+1}dx\\ \implies \frac{1}{2} \int \frac{1}{x-1}dx - \frac{1}{4} \int\frac{2x}{x^2+1}dx + \frac{1}{2} \int\frac{1}{x^2+1}dx $$ After splitting the integral we obtain 3 simpler integrals, the first and third one can be evaluated by using basic integral formulae, the second integral is of the form $\int \frac{f'(x)}{f(x)}dx = \ln|f(x)|+c$

Thus we get,

$$ \frac{1}{2} \ln|x-1| - \frac{1}{4} \ln(x^2+1) + \frac{1}{2} \arctan(x) + C $$ Now substituting back the initial function we get,

$$ \frac{1}{2}\ln\Big|\sqrt{\frac{1}{t^2} - 1} - 1 \Big| - \frac{1}{4} \ln\Big(\frac{1}{t^2}\Big) + \frac{1}{2} \arctan\Big(\sqrt{\frac{1}{t^2} - 1}\Big) + C \\ \implies \frac{1}{2} \ln|t - \sqrt{1-t^2}| + \frac{1}{2} \arccos(t) + C \\= \frac{1}{2} \ln|t - \sqrt{1-t^2}| - \frac{1}{2} \arcsin(t) + C $$ Both answers are correct and only differ by the constant $\frac{\pi}{2}$

Answer 3

$$ I:=\int\frac{\cos xdx}{\sin x-\cos x}=\frac{1}{2}\left(\int\frac{\cos x - \sin x}{\sin x - \cos x} dx + \int\frac{\cos x + \sin x}{\sin x - \cos x} dx\right)\\ =\frac{1}{2}\left(-\int dx + \int\frac{d{(\sin x - \cos x)}}{\sin x - \cos x} \right)\\ =\frac{1}{2}(-x + \ln|\sin x - \cos x|)+c$$

Now, $t=\sin x$

$\cos x=\sqrt{1-t^2}$

$x=\sin^{-1} t$

$$\therefore I = \frac{1}{2}(-\sin^{-1} t + \ln|t - \sqrt{1-t^2}|)+c$$

Answer 4

Substitute $s=\sqrt{\dfrac{1-t}{1+t}} \implies t = \dfrac{1-s^2}{1+s^2}$ :

$$\begin{align*} & \int \frac{dt}{t - \sqrt{1-t^2}} \\ &= \int \frac{\frac{-4s}{\left(1+s^2\right)^2} \, ds}{\frac{1-s^2}{1+s^2} - \sqrt{1-\left(\frac{1-s^2}{1+s^2}\right)^2}} \\ &= 4\int \frac{s}{\left(s^2 + 2s - 1\right) \left(s^2+1\right)} \, ds \\ &= \int \left(\frac{s+1}{s^2+2s-1} - \frac{s-1}{s^2+1}\right) \, ds \end{align*}$$