Question

Integrate with respect to ${}^{\prime }{x}^{\prime }$:
${\sin }^{-1}\sqrt{x}$

Answer 1

$I=\int 1.{\sin }^{-1}\sqrt{x}dx$

$\downarrow$ $\downarrow$

$\left(2\right)$ $\left(1\right)$

$\left\{\begin{array}{c}usingintegrationbypart\\ \left(1\right).\int \left(2\right)-\int (\int (2\left)\right)\times \left(d\right(1\left)\right)\end{array}\right\}$

$I=x.{\sin }^{-1}\sqrt{x}-\int x.\frac{1}{\sqrt{1-(\sqrt{x}{)}^2}}dx$

$I=x.{\sin }^{-1}\sqrt{x}+\int \frac{1-x-1}{\sqrt{1-x}}dx$

$I=x{\sin }^{-1}\sqrt{x}+\int (\frac{1-x}{\sqrt{1-x}}-\frac{1}{\sqrt{1-x}})dx$

$I=x{\sin }^{-1}\sqrt{x}+\int \sqrt{1-x}dx-\int \frac{1}{\sqrt{1-x}}dx$

$\because \int \sqrt{1-x}dx=\sqrt{1^2(\sqrt{x}{)}^2}$

Formula

$\int \sqrt{a^2-x^2}dx=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}$

$\therefore \int \sqrt{1-(\sqrt{x}{)}^{2}dx}=\frac{\sqrt{x}\sqrt{1-(x{)}^2}}{2}+\frac{1}{2}{\sin }^{-1}\frac{\sqrt{x}}{1}$

$\therefore I=xsi{n}^{-1}\sqrt{x}+\frac{\sqrt{x}\sqrt{1-x}}{2}+\frac{1}{2}{\sin }^{-1}\sqrt{x}-\int \frac{1}{\sqrt{1-(\sqrt{x}{)}^2}}dx$

$=x{\sin }^{-1}\sqrt{x}+\frac{\sqrt{x(1-x)}}{2}+\frac{1}{2}{\sin }^{-1}\sqrt{x}-{\sin }^{-1}\sqrt{x}+c$

$=x{\sin }^{-1}\sqrt{x}+\frac{\sqrt{x(1-x)}}{2}-\frac{1}{2}{\sin }^{-1}\sqrt{x}+c$

$=(x-\frac{1}{2}){\sin }^{-1}\sqrt{x}+\frac{1}{2}\sqrt{x(1-x)}+c$