Integral of exp(x)sin(x)sin(x) dx

1. Use integration by parts:

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

   Let $u{\left(x \right)} = \sin^{2}{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = e^{x}$.

   Then $\operatorname{du}{\left(x \right)} = 2 \sin{\left(x \right)} \cos{\left(x \right)}$.

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

   1. The integral of the exponential function is itself.

      $$\int e^{x}\, dx = e^{x}$$

   Now evaluate the sub-integral.

2. Use integration by parts:

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

   Let $u{\left(x \right)} = 2 \sin{\left(x \right)} \cos{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = e^{x}$.

   Then $\operatorname{du}{\left(x \right)} = - 2 \sin^{2}{\left(x \right)} + 2 \cos^{2}{\left(x \right)}$.

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

   1. The integral of the exponential function is itself.

      $$\int e^{x}\, dx = e^{x}$$

   Now evaluate the sub-integral.

3. Rewrite the integrand:

   $$\left(- 2 \sin^{2}{\left(x \right)} + 2 \cos^{2}{\left(x \right)}\right) e^{x} = - 2 e^{x} \sin^{2}{\left(x \right)} + 2 e^{x} \cos^{2}{\left(x \right)}$$

4. Integrate term-by-term:

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

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

      1. Don't know the steps in finding this integral.

         But the integral is

         $$\frac{3 e^{x} \sin^{2}{\left(x \right)}}{5} - \frac{2 e^{x} \sin{\left(x \right)} \cos{\left(x \right)}}{5} + \frac{2 e^{x} \cos^{2}{\left(x \right)}}{5}$$

      So, the result is: $- \frac{6 e^{x} \sin^{2}{\left(x \right)}}{5} + \frac{4 e^{x} \sin{\left(x \right)} \cos{\left(x \right)}}{5} - \frac{4 e^{x} \cos^{2}{\left(x \right)}}{5}$

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

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

      1. Don't know the steps in finding this integral.

         But the integral is

         $$\frac{2 e^{x} \sin^{2}{\left(x \right)}}{5} + \frac{2 e^{x} \sin{\left(x \right)} \cos{\left(x \right)}}{5} + \frac{3 e^{x} \cos^{2}{\left(x \right)}}{5}$$

      So, the result is: $\frac{4 e^{x} \sin^{2}{\left(x \right)}}{5} + \frac{4 e^{x} \sin{\left(x \right)} \cos{\left(x \right)}}{5} + \frac{6 e^{x} \cos^{2}{\left(x \right)}}{5}$

   The result is: $- \frac{2 e^{x} \sin^{2}{\left(x \right)}}{5} + \frac{8 e^{x} \sin{\left(x \right)} \cos{\left(x \right)}}{5} + \frac{2 e^{x} \cos^{2}{\left(x \right)}}{5}$

5. Now simplify:

   $$\frac{\left(- 2 \sin{\left(2 x \right)} - \cos{\left(2 x \right)} + 5\right) e^{x}}{10}$$

6. Add the constant of integration:

   $$\frac{\left(- 2 \sin{\left(2 x \right)} - \cos{\left(2 x \right)} + 5\right) e^{x}}{10}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(- 2 \sin{\left(2 x \right)} - \cos{\left(2 x \right)} + 5\right) e^{x}}{10}+ \mathrm{constant}$$