# How do you integrate int sqrt(arcsinx/(1-x^2) using substitution?

Oct 9, 2016

Not only does this integral needs a standard Trig substitution, but also an inverse trig identity not commonly used.

#### Explanation:

int(sqrt((sin^-1)/(1-x^2))) dx

$x = \cos u \Rightarrow \mathrm{dx} = - \sin u \mathrm{du}$
${\cos}^{1} x = u$

${\sin}^{-} 1 + {\cos}^{-} 1 = \frac{\pi}{2}$

${\sin}^{-} 1 x = \left(\frac{\pi}{2}\right) - {\cos}^{-} 1 u$
${\sin}^{-} 1 x = \frac{\pi}{2} - u$

substituting into the integral

$\int \left(\sqrt{\frac{\frac{\pi}{2} - u}{1 - {\cos}^{2} u}}\right) \times \left(- \sin u \mathrm{du}\right)$

$- \int \left({\left(\frac{\pi}{2} - u\right)}^{\frac{1}{2}} / \sqrt{{\sin}^{2} u}\right) \sin u \mathrm{du}$

$- \int \left({\left(\frac{\pi}{2} - u\right)}^{\frac{1}{2}} / \left(\sin u\right)\right) \sin u \mathrm{du}$

$\sin u$ is cancelled from numerator and denominator

$- \int {\left(\frac{\pi}{2} - u\right)}^{\frac{1}{2}} \mathrm{du}$

integrating by inspection

$= \frac{2}{3} {\left(\frac{\pi}{2} - u\right)}^{\frac{3}{2}} + C$

substitute back for $u$

$= \frac{2}{3} {\left(\frac{\pi}{2} - {\cos}^{-} 1 x\right)}^{\frac{3}{2}} + C$

$= \frac{2}{3} {\left({\sin}^{-} 1 x\right)}^{\frac{3}{2}} + C$

Oct 9, 2016

$\frac{2}{3} {\left(\arcsin x\right)}^{\frac{3}{2}} + C$

#### Explanation:

$\int \sqrt{\arcsin \frac{x}{1 - {x}^{2}}} \mathrm{dx}$

Note that the square root can be split up:

$= \int \sqrt{\arcsin} \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

Furthermore, note that we have an $\arcsin$ function and its derivative present:

$= \int {\left(\arcsin x\right)}^{\frac{1}{2}} \left(\frac{1}{\sqrt{1 - {x}^{2}}}\right) \mathrm{dx}$

Using substitution, where $u = \arcsin x$ and $\mathrm{du} = \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$:

$= \int {u}^{\frac{1}{2}} \mathrm{du}$

Using the typical power rule for integration:

$= {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + C = \frac{2}{3} {u}^{\frac{3}{2}} + C = \frac{2}{3} {\left(\arcsin x\right)}^{\frac{3}{2}} + C$