Integral of 4sin^2t dx

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

   $$\int 4 \sin^{2}{\left(t \right)}\, dt = 4 \int \sin^{2}{\left(t \right)}\, dt$$

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

         $$\int \frac{1}{2}\, dt = \frac{t}{2}$$

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

         $$\int \left(- \frac{\cos{\left(2 t \right)}}{2}\right)\, dt = - \frac{\int \cos{\left(2 t \right)}\, dt}{2}$$

         1. Let $u = 2 t$.

            Then let $du = 2 dt$ 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:

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

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

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

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

2. Add the constant of integration:

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

The answer is: