Integral of arcsen(x) dx

1. Use integration by parts:

   $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

   Let $u{\left(x \right)} = \operatorname{asin}{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

   Then $\operatorname{du}{\left(x \right)} = \frac{1}{\sqrt{1 - x^{2}}}$.

   To find $v{\left(x \right)}$:

   1. The integral of a constant is the constant times the variable of integration:

      $$\int 1\, dx = x$$

   Now evaluate the sub-integral.

2. Let $u = 1 - x^{2}$.

   Then let $du = - 2 x dx$ and substitute $- \frac{du}{2}$:

   $$\int \left(- \frac{1}{2 \sqrt{u}}\right)\, du$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \frac{1}{\sqrt{u}}\, du = - \frac{\int \frac{1}{\sqrt{u}}\, du}{2}$$

      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int \frac{1}{\sqrt{u}}\, du = 2 \sqrt{u}$$

      So, the result is: $- \sqrt{u}$

   Now substitute $u$ back in:

   $$- \sqrt{1 - x^{2}}$$

3. Add the constant of integration:

   $$x \operatorname{asin}{\left(x \right)} + \sqrt{1 - x^{2}}+ \mathrm{constant}$$

The answer is:

$$x \operatorname{asin}{\left(x \right)} + \sqrt{1 - x^{2}}+ \mathrm{constant}$$