Integration of $\int\frac{x-1}{\sqrt{1+\sqrt{1-x^2}}}dx$

Question

Integral of $\int\frac{x-1}{\sqrt{1+\sqrt{1-x^2}}}dx$

This is the answer, but I'm not sure how to get to it...

http://www.wolframalpha.com/input/?i=Integral+of+(x-1)%2F((sqrt(1%2Bsqrt(1-x%5E2))))dx

Answer 1

HINT: If you substitute $x=\sin 2\theta$, then you will get $$\sqrt2\int \frac{(\sin 2\theta-1)\cos 2\theta}{\cos\theta}\,d\theta.$$ Note that we have to use the formula $\cos 2\theta=2\cos^2\theta-1$. Now, use the double angle formulas to rewrite $\sin 2\theta$ and $\cos 2\theta$, and you should be good to go. You'll get basic integrals (including $\int\sec\theta\,d\theta$).

Answer 2

$$I=\int\frac{x-1}{\sqrt{1+\sqrt{1-x^2}}}dx$$ Start from substitution $t^2=1-x^2,\quad x=\sqrt{1-t^2},\quad dx=-\frac{t}{\sqrt{1-t^2}}dt$ $$ I=\int\frac{t(1-\sqrt{1-t^2})}{\sqrt{1+t}\sqrt{1-t^2}}dt=\int\frac{t}{\sqrt{1+t}\sqrt{1-t^2}}dt-\int\frac{t}{\sqrt{1+t}}dt=I_1-I_2+const $$ $I_2$ is trivial and equal to $\frac{2}{3}(t-2)\sqrt{t+1}$, $I_1 $ requires further transformation $$ I_1=\int\frac{t}{\sqrt{1+t}\sqrt{1-t^2}}dt=\int\frac{t}{(1+t)\sqrt{1-t}}dt\\ I_1=\int\frac{1}{\sqrt{1-t}}dt-\int\frac{1}{(1+t)\sqrt{1-t}}dt $$ First integral is trivial, second is also can be considered as a table one (alternatively use substitution $z^2=1-t$). $$ I_1=(-2\sqrt{1-t})-\left(-\sqrt{2}\tanh^{-1}\left(\frac{1-t}{2}\right)\right) $$ Hence, $$ I=-2\sqrt{1-t}+\left(\sqrt{2}\tanh^{-1}\left(\frac{1-t}{2}\right)\right)-\frac{2}{3}(t-2)\sqrt{1+t}+const $$ Or in initial variables $$ I=-2\sqrt{1-\sqrt{1-x^2}}+\left(\sqrt{2}\tanh^{-1}\left(\frac{1-\sqrt{1-x^2}}{2}\right)\right)-\frac{2}{3}(\sqrt{1-x^2}-2)\sqrt{1+\sqrt{1-x^2}}+const $$

Answer 3

Boy can you learn alot from this question. Using a "standard technique" one arrives at the substitution:

$$x=\frac{4s(s^2-1)}{(s^2+1)^2}$$ This will make the integral into a rational function of $s$. This substitution is invertible via

$$s=\frac{\sqrt{2-2\sqrt{1-x^2}-x^2}-\sqrt{2}\sqrt{1+\sqrt{1-x^2}}}{x}$$

(and yes, there is a logic to this).

To see that the integral is rational we need that $\sqrt{1+\sqrt{1-x^2}}$ is rational. One can calculate that $$\sqrt{1-x^2}=\frac{s^4-6s^2+1}{(s^2+1)^2}$$ and that $$\sqrt{1+\sqrt{1-x^2}}=\frac{2\sqrt{2}s}{s^2+1}$$

The rest is a standard substitution, to a rational function and then partial fractions can be applied. The denominator is actually quite nice, the numerator is a messy polynomial, but that is not too serious, then just substitute to get back to $x$.