Evaluating $\displaystyle{\int{\dfrac{\arcsin{x}}{\sqrt{1 + x}}}}$ with minimum substitutions.

Question

$$\int{\dfrac{\arcsin{x}}{\sqrt{1 + x}}} \mathrm{dx}$$

This is a Q&A post. To clarify the title better, the challenge of the original question is to evaluate the integral using as less as possible, or even no $u$-substitutions.

My working is provided in the answer below. I am welcome to feedback and simpler methods that can complete the challenge better.

Answer 1

I am getting a slightly different result. Applying integration by parts: $$2\int{\dfrac{\arcsin{x}}{2\sqrt{1 + x}}} \space \mathrm{dx} = 2\sqrt{1 + x}\arcsin{x}- 2\int{\dfrac{\sqrt{1+x}}{\sqrt{1 - x^2}}} \mathrm{dx}= \\ 2\sqrt{1 + x}\arcsin{x}- 2\int{\dfrac{1}{\sqrt{1 - x}}} \mathrm{dx}= \\ 2\sqrt{1 + x}\arcsin{x} + 4\sqrt{1 - x} +C\\ $$

Answer 2

$$\int{\dfrac{\arcsin{x}}{\sqrt{1 + x}}} \space \mathrm{dx} = \int{\dfrac{\arcsin{(x)}\sqrt{1-x}}{\sqrt{1 - x^2}}} \mathrm{dx} \\ \xrightarrow{\large{\begin{cases}t \space = \space \arcsin{x} \\ dt \space = \space \tfrac{1}{ \sqrt{1-x^2}} \space dx\end{cases}}} \int{t\sqrt{1 - \sin{t}}} \space \mathrm{dt}\\ \xrightarrow{\large{\text{IBP}}} t\int{\sqrt{1 - \sin{t}}} \space \mathrm{dt} - \int\int{\sqrt{1 - \sin{t}}} \space \mathrm{dt} \space \mathrm{dt}\\$$

$$\sqrt{1 - \sin{t}} = \sqrt{1 - \cos{\left(\dfrac{\pi}{2} - t \right)}} = \sqrt{1 - \left( 1 - 2\sin^2{\left(\dfrac{\pi}{4} - \dfrac{t}{2} \right)} \right)} = \sqrt{2}\left|\sin{\left(\dfrac{\pi}{4} - \dfrac{t}{2} \right)}\right| = \sqrt{2}\sin{\left(\dfrac{\pi}{4} - \dfrac{t}{2} \right)}, \\ \text{ because, } 0 < \sin{\left(\dfrac{\pi}{4} - \dfrac{t}{2} \right)} < 1 \text{ for } \dfrac{-\pi}{2} < t < \dfrac{\pi}{2}$$

$$\to \sqrt{2}t\int{\sin{\left(\dfrac{\pi}{4} - \dfrac{t}{2} \right)} } \space \mathrm{dt} - \sqrt{2}\int\int{\sin{\left(\dfrac{\pi}{4} - \dfrac{t}{2} \right)} } \space \mathrm{dt} \space \mathrm{dt} \\ \quad \\ = 4\sqrt{2}\cos{\left(\dfrac{\pi}{4} + \dfrac{t}{2} \right)} + 2t\sqrt{2}\sin{\left(\dfrac{\pi}{4} + \dfrac{t}{2} \right)} + C \\ \quad \\ = 4\cos{\dfrac{t}{2}} - 4\sin{\dfrac{t}{2}} + 2t \cos{\dfrac{t}{2}} + 2t\sin{\dfrac{t}{2}} + C$$

$\text{Using, } \\ \cos{\dfrac{\arcsin{x}}{2}} = \sqrt{\dfrac{1 + \cos{\arcsin{x}}}{2}} = \dfrac{\sqrt{1 - x}}{2} + \dfrac{\sqrt{1 + x}}{2} \\ \sin{\dfrac{\arcsin{x}}{2}} = \sqrt{\dfrac{1 - \cos{\arcsin{x}}}{2}} = -\dfrac{\sqrt{1 - x}}{2} + \dfrac{\sqrt{1 + x}}{2}$.

The final answer would be,

$$4\cdot\sqrt{1 - x} + 2 \cdot \arcsin{x} \cdot \sqrt{1 + x} + C$$

Currently, I have employed $1$ $u-$sub. Feel welcome to post other elementary methods, (especially those without u-sub completely).

Answer 3

$$ \begin{align}\int\frac{\arcsin(x)}{\sqrt{1+x}}\text{d}x &=2\int\arcsin(x)\text{d}\sqrt{1+x},\ \ u=\sqrt{1+x}\Rightarrow x=u^2-1\\&=2\int\arcsin(u^2-1)\text{d}u,\ \ \text{IBP}\Rightarrow\text{d}[\arcsin(u^2-1)]=\frac{2u}{\sqrt{2-u^2}}\\&=2u\arcsin(u^2-1)-2\int\frac{2u}{\sqrt{2-u^2}}\text{d}u\\&=2u\arcsin(u^2-1)-2\int\frac{1}{\sqrt{2-u^2}}\text{d}u^2,\ \ t=u^2\\&=2u\arcsin(u^2-1)-2\int\frac{1}{\sqrt{2-t}}\text{d}t\\&=2u\arcsin(u^2-1)+2\int\frac{1}{\sqrt{2-t}}\text{d}(2-t)\\&=2u\arcsin(u^2-1)+4\sqrt{2-t},\ \ t=u^2=1+x\\&=2\sqrt{1+x}\arcsin(x)+4\sqrt{1-x}+C\end{align}$$