Integral of arcsin3x dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 3 x$.

      Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

      $$\int \frac{\operatorname{asin}{\left(u \right)}}{3}\, du$$

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

         $$\int \operatorname{asin}{\left(u \right)}\, du = \frac{\int \operatorname{asin}{\left(u \right)}\, du}{3}$$

         1. Use integration by parts:

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

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

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

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

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

               $$\int 1\, du = u$$

            Now evaluate the sub-integral.

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

            Then let $du = - 2 u du$ 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:

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

         So, the result is: $\frac{u \operatorname{asin}{\left(u \right)}}{3} + \frac{\sqrt{1 - u^{2}}}{3}$

      Now substitute $u$ back in:

      $$x \operatorname{asin}{\left(3 x \right)} + \frac{\sqrt{1 - 9 x^{2}}}{3}$$

   Method #2

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = \operatorname{asin}{\left(3 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

      Then $\operatorname{du}{\left(x \right)} = \frac{3}{\sqrt{1 - 9 x^{2}}}$.

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

      $$\int \frac{3 x}{\sqrt{1 - 9 x^{2}}}\, dx = 3 \int \frac{x}{\sqrt{1 - 9 x^{2}}}\, dx$$

      1. Let $u = 1 - 9 x^{2}$.

         Then let $du = - 18 x dx$ and substitute $- \frac{du}{18}$:

         $$\int \left(- \frac{1}{18 \sqrt{u}}\right)\, du$$

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

            $$\int \frac{1}{\sqrt{u}}\, du = - \frac{\int \frac{1}{\sqrt{u}}\, du}{18}$$

            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: $- \frac{\sqrt{u}}{9}$

         Now substitute $u$ back in:

         $$- \frac{\sqrt{1 - 9 x^{2}}}{9}$$

      So, the result is: $- \frac{\sqrt{1 - 9 x^{2}}}{3}$

2. Add the constant of integration:

   $$x \operatorname{asin}{\left(3 x \right)} + \frac{\sqrt{1 - 9 x^{2}}}{3}+ \mathrm{constant}$$

The answer is:

$$x \operatorname{asin}{\left(3 x \right)} + \frac{\sqrt{1 - 9 x^{2}}}{3}+ \mathrm{constant}$$