Integral of sin(3x)sinxsinx dx

1. Rewrite the integrand:

   $$\sin{\left(x \right)} \sin{\left(3 x \right)} \sin{\left(x \right)} = - 4 \sin^{5}{\left(x \right)} + 3 \sin^{3}{\left(x \right)}$$

2. Integrate term-by-term:

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

      $$\int \left(- 4 \sin^{5}{\left(x \right)}\right)\, dx = - 4 \int \sin^{5}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\sin^{5}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)}$$

      2. There are multiple ways to do this integral.

         Method #1

         1. Rewrite the integrand:

            $$\left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} = \sin{\left(x \right)} \cos^{4}{\left(x \right)} - 2 \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}$$

         2. Integrate term-by-term:

            1. Let $u = \cos{\left(x \right)}$.

               Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

               $$\int \left(- u^{4}\right)\, du$$

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

                  $$\int u^{4}\, du = - \int u^{4}\, du$$

                  1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                     $$\int u^{4}\, du = \frac{u^{5}}{5}$$

                  So, the result is: $- \frac{u^{5}}{5}$

               Now substitute $u$ back in:

               $$- \frac{\cos^{5}{\left(x \right)}}{5}$$

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

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

               1. Let $u = \cos{\left(x \right)}$.

                  Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

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

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

                     $$\int u^{2}\, du = - \int u^{2}\, du$$

                     1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                        $$\int u^{2}\, du = \frac{u^{3}}{3}$$

                     So, the result is: $- \frac{u^{3}}{3}$

                  Now substitute $u$ back in:

                  $$- \frac{\cos^{3}{\left(x \right)}}{3}$$

               So, the result is: $\frac{2 \cos^{3}{\left(x \right)}}{3}$

            1. The integral of sine is negative cosine:

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

            The result is: $- \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$

         Method #2

         1. Rewrite the integrand:

            $$\left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} = \sin{\left(x \right)} \cos^{4}{\left(x \right)} - 2 \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}$$

         2. Integrate term-by-term:

            1. Let $u = \cos{\left(x \right)}$.

               Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

               $$\int \left(- u^{4}\right)\, du$$

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

                  $$\int u^{4}\, du = - \int u^{4}\, du$$

                  1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                     $$\int u^{4}\, du = \frac{u^{5}}{5}$$

                  So, the result is: $- \frac{u^{5}}{5}$

               Now substitute $u$ back in:

               $$- \frac{\cos^{5}{\left(x \right)}}{5}$$

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

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

               1. Let $u = \cos{\left(x \right)}$.

                  Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$:

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

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

                     $$\int u^{2}\, du = - \int u^{2}\, du$$

                     1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

                        $$\int u^{2}\, du = \frac{u^{3}}{3}$$

                     So, the result is: $- \frac{u^{3}}{3}$

                  Now substitute $u$ back in:

                  $$- \frac{\cos^{3}{\left(x \right)}}{3}$$

               So, the result is: $\frac{2 \cos^{3}{\left(x \right)}}{3}$

            1. The integral of sine is negative cosine:

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

            The result is: $- \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$

      So, the result is: $\frac{4 \cos^{5}{\left(x \right)}}{5} - \frac{8 \cos^{3}{\left(x \right)}}{3} + 4 \cos{\left(x \right)}$

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

      $$\int 3 \sin^{3}{\left(x \right)}\, dx = 3 \int \sin^{3}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\sin^{3}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}$$

      2. Let $u = \cos{\left(x \right)}$.

         Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$:

         $$\int \left(u^{2} - 1\right)\, du$$

         1. Integrate term-by-term:

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int u^{2}\, du = \frac{u^{3}}{3}$$

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

               $$\int \left(-1\right)\, du = - u$$

            The result is: $\frac{u^{3}}{3} - u$

         Now substitute $u$ back in:

         $$\frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$$

      So, the result is: $\cos^{3}{\left(x \right)} - 3 \cos{\left(x \right)}$

   The result is: $\frac{4 \cos^{5}{\left(x \right)}}{5} - \frac{5 \cos^{3}{\left(x \right)}}{3} + \cos{\left(x \right)}$

3. Now simplify:

   $$\frac{\left(12 \sin^{4}{\left(x \right)} + \sin^{2}{\left(x \right)} + 2\right) \cos{\left(x \right)}}{15}$$

4. Add the constant of integration:

   $$\frac{\left(12 \sin^{4}{\left(x \right)} + \sin^{2}{\left(x \right)} + 2\right) \cos{\left(x \right)}}{15}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(12 \sin^{4}{\left(x \right)} + \sin^{2}{\left(x \right)} + 2\right) \cos{\left(x \right)}}{15}+ \mathrm{constant}$$