Integral of 2sin²x dx

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

   $$\int 2 \sin^{2}{\left(x \right)}\, dx = 2 \int \sin^{2}{\left(x \right)}\, dx$$

   1. Rewrite the integrand:

      $$\sin^{2}{\left(x \right)} = \frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}$$

   2. Integrate term-by-term:

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

         $$\int \frac{1}{2}\, dx = \frac{x}{2}$$

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

         $$\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}$$

         1. Let $u = 2 x$.

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

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

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

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

               1. The integral of cosine is sine:

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

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

            Now substitute $u$ back in:

            $$\frac{\sin{\left(2 x \right)}}{2}$$

         So, the result is: $- \frac{\sin{\left(2 x \right)}}{4}$

      The result is: $\frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}$

   So, the result is: $x - \frac{\sin{\left(2 x \right)}}{2}$

2. Add the constant of integration:

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

The answer is: