Integral of sin(sqrt(x)) dx

1. Let $u = \sqrt{x}$.

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

   $$\int 2 u \sin{\left(u \right)}\, du$$

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

      $$\int u \sin{\left(u \right)}\, du = 2 \int u \sin{\left(u \right)}\, du$$

      1. Use integration by parts:

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

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

         Then $\operatorname{du}{\left(u \right)} = 1$.

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

         1. The integral of sine is negative cosine:

            $$\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$$

         Now evaluate the sub-integral.

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

         $$\int \left(- \cos{\left(u \right)}\right)\, du = - \int \cos{\left(u \right)}\, du$$

         1. The integral of cosine is sine:

            $$\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$$

         So, the result is: $- \sin{\left(u \right)}$

      So, the result is: $- 2 u \cos{\left(u \right)} + 2 \sin{\left(u \right)}$

   Now substitute $u$ back in:

   $$- 2 \sqrt{x} \cos{\left(\sqrt{x} \right)} + 2 \sin{\left(\sqrt{x} \right)}$$

2. Add the constant of integration:

   $$- 2 \sqrt{x} \cos{\left(\sqrt{x} \right)} + 2 \sin{\left(\sqrt{x} \right)}+ \mathrm{constant}$$

The answer is:

$$- 2 \sqrt{x} \cos{\left(\sqrt{x} \right)} + 2 \sin{\left(\sqrt{x} \right)}+ \mathrm{constant}$$