# How do you findintarcsinx/sqrt(1-x^2)dx?

Jul 22, 2018

$\frac{1}{2} {\left(\arcsin x\right)}^{2} + C$.

#### Explanation:

If we subst. $\arcsin x = t , \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = \mathrm{dt}$.

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

$= t \mathrm{dt}$,

$= \frac{1}{2} {t}^{2}$,

$= \frac{1}{2} {\left(\arcsin x\right)}^{2} + C$.

Jul 22, 2018

The answer is $= {\left(\arcsin x\right)}^{2} / 2 + C$

#### Explanation:

Perform this integral by substitution

Let $u = \arcsin x$, $\implies$, $\mathrm{du} = \frac{\mathrm{dx}}{\sqrt{1 - {x}^{2}}}$

The integral is

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

$= \int u \mathrm{du}$

$= {u}^{2} / 2$

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

Jul 22, 2018

$\int \frac{a r c \sin x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = {\left(\arcsin x\right)}^{2} / 2 + c$

#### Explanation:

Here,

$I = \int \frac{a r c \sin x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = \int a r c \sin x \cdot \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

Subst. color(blue)(arc sinx=u=>1/sqrt(1-x^2)dx=du

$\therefore I = \int u \mathrm{du}$

$\implies I = {u}^{2} / 2 + c$

Subst. back color(blue)(u=arc sinx

$\therefore I = {\left(\arcsin x\right)}^{2} / 2 + c$