$\int \arcsin\sqrt{1-x^2} dx$ - how to evaluate?

Question

I've been trying to solve this integral $$ \int \arcsin\sqrt{1-x^2}\, dx, $$
but got an expression with an absolute value of $x$. How can I solve this integral?

Answer 1

Integration by parts.

$$f(x) = \arcsin(\sqrt{1-x^2})$$ $$g'(x) = 1$$

By the rule you get

$$\int 1 \cdot \arcsin\sqrt{1-x^2}\ \text{d}x = x\arcsin\sqrt{1-x^2} + \int x\cdot \frac{x}{1-x^2}\ \text{d}x$$

The latter integral is rather simple, so you can proceed from here. ($x > 0$)

Answer 2

Let's do the $x\ge0$ case first. Integrate by parts with $u=\arcsin\sqrt{1-x^2},\,v=x$ so the integral is$$\begin{align}x\arcsin\sqrt{1-x^2}-\int x\frac{d}{dx}\left(\arcsin\sqrt{1-x^2}\right)dx&=x\arcsin\sqrt{1-x^2}+\int\frac{xdx}{\sqrt{1-x^2}}\\&=x\arcsin\sqrt{1-x^2}-\sqrt{1-x^2}+C.\end{align}$$Now for the general problem. Since the original integrand is even, the antiderivative is an odd function $+C$. So the general antiderivative is$$x\arcsin\sqrt{1-x^2}-\frac{|x|}{x}\sqrt{1-x^2}+C.$$The fraction is also denoted $\operatorname{sgn}x$ or $\operatorname{sign}x$.

Answer 3

Since you got an absolute value, my guess is that what you did was $x=\cos y$ and $\mathrm dx=-\sin(y)\,\mathrm dy$. So, you got$$\int\arcsin\left(\sqrt{1-\cos^2y}\right)\,\mathrm dy=-\int\arcsin\left(|\sin y|\right)\sin(y)\,\mathrm dy.$$Don't worry about it. Deal with it as if $\sqrt{1-\cos^2y}=\sin y$ and therefore $\arcsin\left(\sqrt{1-\cos^2y}\right)=y$. All will be fine.

Answer 4

Let $\arcsin\sqrt{1-x^2}=y$

As $\sqrt{1-x^2}\ge0, 0\le y\le\dfrac\pi2$

$x^2=\cos^2y$

As $\cos y\ge0$

if $x\ge0, x=\cos y, y=\arccos(x)=\dfrac\pi2-\arcsin x$

if $x<0,x=-\cos y, y=\arccos(-x)$

Using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?, $$y=\pi-\arccos x=\pi-\left(\dfrac\pi2-\arcsin x\right)$$

Alternatively, $$y=\dfrac\pi2-\arcsin(-x)=\dfrac\pi2+\arcsin x$$

Now integrate by parts

$$\int\arcsin x\ dx=\arcsin x\int\ dx-\int\left(\dfrac{d(\arcsin x)}{dx}\int\ dx\right)dx$$