$$\int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)dx$$
I'm trying to solve this integral from GN Berman's Problems on a course of mathematical analysis (Question number 1845)
I tried substituting $x$ for $t^2$: $$2\int t\arcsin\left(\frac{t}{\sqrt{1-t^2}}\right) dt$$ And then evaluating it by parts, which made it very complicated.
So how do I go about solving this?
Integrate by parts
\begin{align} &\int \arcsin\sqrt{\frac{x}{1-x}}\ d(x-1)\\ =& \ (x-1) \arcsin\sqrt{\frac{x}{1-x}} +\frac12\int \frac1{\sqrt{x(1-2x)}}dx \\ =& \ (x-1) \arcsin\sqrt{\frac{x}{1-x}} +\frac1{\sqrt2}\sin^{-1}\sqrt{2x}+C \end{align}
Here is a nice solution that utilizes symmetry instead of $u$ susbsitution. Notice that the domain of this function is at most $\left[0,\frac{1}{2}\right]$, so for the moment let's define
$$F(x) \equiv \int_0^x \arcsin\left(\sqrt{\frac{t}{1-t}}\right)\:dt$$
We could do this integral directly, or since the integrand has a fixed point at $0$, note that the integral is the same as the area of the box $x \cdot \arcsin\left(\sqrt{\frac{x}{1-x}}\right)$ minus the integral of the inverse function:
$$F(x) = x \arcsin\left(\sqrt{\frac{x}{1-x}}\right) - \int_0^{\arcsin\left(\sqrt{\frac{x}{1-x}}\right)} 1 - \frac{1}{1+\sin^2 y}\:dy$$
$$= (x-1) \arcsin\left(\sqrt{\frac{x}{1-x}}\right) + \int_0^{\arcsin\left(\sqrt{\frac{x}{1-x}}\right)}\frac{1}{\cos^2y+2\sin^2 y}\:dy$$
The integral reduces to
$$\int_0^{\arcsin\left(\sqrt{\frac{x}{1-x}}\right)}\frac{\sec^2y}{1+2\tan^2 y}\:dy = \frac{1}{\sqrt{2}}\arctan\left(\sqrt{2}\tan y\right)\Biggr|_0^{\arcsin\left(\sqrt{\frac{x}{1-x}}\right)}$$
$$ = \frac{1}{\sqrt{2}}\arctan\left(\sqrt{\frac{2x}{1-2x}}\right)$$
which gives us a final answer of
$$F(x) = (x-1) \arcsin\left(\sqrt{\frac{x}{1-x}}\right)+\frac{1}{\sqrt{2}}\arctan\left(\sqrt{\frac{2x}{1-2x}}\right)$$
making the most general antiderivative
$$\int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)\:dx = (x-1) \arcsin\left(\sqrt{\frac{x}{1-x}}\right)+\frac{1}{\sqrt{2}}\arctan\left(\sqrt{\frac{2x}{1-2x}}\right) + C$$
This solution has the benefit of not having complicated substitutions to undo, like the other solutions have, but we got lucky since this technique will not always work (what are the odds of the inverse function being more easily integrable than the original?)
$$I=\int \sin^{-1} \sqrt{\frac{x}{1-x}} dx$$ Let $$\frac{x}{1-x}=t^2 \implies x=\frac{t^2}{1+t^2} \implies dx=\frac{2t}{(1+t^2)^2}$$ Then $$I=\int\frac{2t \sin^{-1}t}{(1+t^2)^2} ~dt $$ Integrate it by parts taking $\sin^{-1} t$ as the first function. Then $$I=\sin^{-1} t \int \frac{ 2t dt}{(1+t^2)^2}=\frac{-\sin^{-1}t}{1+t^2}+\int \frac{dt}{(1+t^2)\sqrt{1-t^2}}=f(t)+I_1$$ Let $t=1/u \implies du=- dt/t^2$ and then let $v=u^2$, then $$I_1=-\int \frac{u du}{(1+u^2)\sqrt{u^2-1}}=-\int \frac{v dv}{2(1+v)\sqrt{v-1}}+C$$
Next take $v=w^2 \implies dv=2wdw$, then $$I_1=-\int \frac{dw}{2+w^2}=-\frac{1}{\sqrt{2}} \tan^{-1} \frac{w}{\sqrt{2}}=-\frac{1}{\sqrt{2}}\tan^{-1} \frac{1-2x}{2x}$$ Finally, $$I=(x-1)\sin^{-1} \sqrt{\frac{x}{1-x}}-\frac{1}{\sqrt{2}}\tan^{-1} \frac{1-2x}{2x}$$
Substitute
$$u = \sqrt{\frac{x}{1-x}} \implies \frac{du}{dx} = \frac{\sqrt{x}}{(1-x)^{\frac{3}{2}}} + \frac{1}{2\sqrt{1-x}\sqrt{x}}$$
Then the integral become:
$$2 \int \frac{u\arcsin(u)}{(u^2 + 1)^2} \,du$$
Integrating by parts $\left(f = \arcsin(u), g'= \frac{u}{(u^2 + 1)^2} \implies f'= \frac{1}{\sqrt{1 - u^2}}, g = -\frac{1}{2(u^2 + 1)}\right)$:
$$2 \int \frac{u\arcsin(u)}{(u^2 + 1)^2} \,du = -\frac{\arcsin u}{2(u^2+1)} - \int \frac{1}{2\sqrt{1-u^2}(u^2+1)} \,du \text{ (1)}$$
We will now solve this integral first:
$$\int \frac{1}{2\sqrt{1-u^2}(u^2+1)} \,du \text{ (2)}$$
Substitute $v = \arcsin(u) \implies du = \cos(v) \,dv$, then:
$$(2) \implies \int \frac{\cos(v)}{2\sqrt{1-\sin(v)^2}(\sin(v)^2+1)} \,dv$$ $$= \frac{1}{2} \int \frac{1}{\sin(v)^2+1} \,dv$$ $$= \frac{1}{2} \int \sec^2(v) \cdot \frac{1}{2\tan(v)^2+1} \,dv \text{ (3)}$$
Substitute $w = \sqrt{2}\tan(v) \implies \frac{dw}{dv} = \sqrt{2}\sec^2(v)$, then:
$$(3) \implies \frac{1}{2\sqrt{2}} \int \frac{1}{w^2+1} \,dw = \frac{1}{2\sqrt{2}} \arctan(w)$$
Undo substitution, we get:
$$\int \frac{1}{2\sqrt{1-u^2}(u^2+1)} \,du = \frac{\arctan(\frac{\sqrt{2}u}{\sqrt{1-u^2}})}{2\sqrt{2}}$$
Plug it back into $(1)$ and undo substitution to get the answer.
This is what you should get. I am very sorry I thought in the denominator its not arcSin, my apologies!!
Please go through this, and let me know if you have concerns or questions.
