Integral of cos(5x)*(cos(2x))^2 dx

1. Rewrite the integrand:

   $$\cos^{2}{\left(2 x \right)} \cos{\left(5 x \right)} = 64 \cos^{9}{\left(x \right)} - 144 \cos^{7}{\left(x \right)} + 116 \cos^{5}{\left(x \right)} - 40 \cos^{3}{\left(x \right)} + 5 \cos{\left(x \right)}$$

2. Integrate term-by-term:

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

      $$\int 64 \cos^{9}{\left(x \right)}\, dx = 64 \int \cos^{9}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\cos^{9}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{4} \cos{\left(x \right)}$$

      2. There are multiple ways to do this integral.

         Method #1

         1. Rewrite the integrand:

            $$\left(1 - \sin^{2}{\left(x \right)}\right)^{4} \cos{\left(x \right)} = \sin^{8}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)} + 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$$

         2. Integrate term-by-term:

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

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

               $$\int u^{8}\, du$$

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

                  $$\int u^{8}\, du = \frac{u^{9}}{9}$$

               Now substitute $u$ back in:

               $$\frac{\sin^{9}{\left(x \right)}}{9}$$

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

               $$\int \left(- 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

                  $$\int u^{6}\, du$$

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

                     $$\int u^{6}\, du = \frac{u^{7}}{7}$$

                  Now substitute $u$ back in:

                  $$\frac{\sin^{7}{\left(x \right)}}{7}$$

               So, the result is: $- \frac{4 \sin^{7}{\left(x \right)}}{7}$

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

               $$\int 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 6 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

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

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

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

                  Now substitute $u$ back in:

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

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

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

               $$\int \left(- 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

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

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

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

                  Now substitute $u$ back in:

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

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

            1. The integral of cosine is sine:

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

            The result is: $\frac{\sin^{9}{\left(x \right)}}{9} - \frac{4 \sin^{7}{\left(x \right)}}{7} + \frac{6 \sin^{5}{\left(x \right)}}{5} - \frac{4 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$

         Method #2

         1. Rewrite the integrand:

            $$\left(1 - \sin^{2}{\left(x \right)}\right)^{4} \cos{\left(x \right)} = \sin^{8}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)} + 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$$

         2. Integrate term-by-term:

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

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

               $$\int u^{8}\, du$$

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

                  $$\int u^{8}\, du = \frac{u^{9}}{9}$$

               Now substitute $u$ back in:

               $$\frac{\sin^{9}{\left(x \right)}}{9}$$

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

               $$\int \left(- 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

                  $$\int u^{6}\, du$$

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

                     $$\int u^{6}\, du = \frac{u^{7}}{7}$$

                  Now substitute $u$ back in:

                  $$\frac{\sin^{7}{\left(x \right)}}{7}$$

               So, the result is: $- \frac{4 \sin^{7}{\left(x \right)}}{7}$

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

               $$\int 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 6 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

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

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

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

                  Now substitute $u$ back in:

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

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

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

               $$\int \left(- 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

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

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

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

                  Now substitute $u$ back in:

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

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

            1. The integral of cosine is sine:

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

            The result is: $\frac{\sin^{9}{\left(x \right)}}{9} - \frac{4 \sin^{7}{\left(x \right)}}{7} + \frac{6 \sin^{5}{\left(x \right)}}{5} - \frac{4 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$

      So, the result is: $\frac{64 \sin^{9}{\left(x \right)}}{9} - \frac{256 \sin^{7}{\left(x \right)}}{7} + \frac{384 \sin^{5}{\left(x \right)}}{5} - \frac{256 \sin^{3}{\left(x \right)}}{3} + 64 \sin{\left(x \right)}$

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

      $$\int \left(- 144 \cos^{7}{\left(x \right)}\right)\, dx = - 144 \int \cos^{7}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

         $$\cos^{7}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{3} \cos{\left(x \right)}$$

      2. Rewrite the integrand:

         $$\left(1 - \sin^{2}{\left(x \right)}\right)^{3} \cos{\left(x \right)} = - \sin^{6}{\left(x \right)} \cos{\left(x \right)} + 3 \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 3 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$$

      3. Integrate term-by-term:

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

            $$\int \left(- \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

               $$\int u^{6}\, du$$

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

                  $$\int u^{6}\, du = \frac{u^{7}}{7}$$

               Now substitute $u$ back in:

               $$\frac{\sin^{7}{\left(x \right)}}{7}$$

            So, the result is: $- \frac{\sin^{7}{\left(x \right)}}{7}$

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

            $$\int 3 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 3 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

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

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

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

               Now substitute $u$ back in:

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

            So, the result is: $\frac{3 \sin^{5}{\left(x \right)}}{5}$

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

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

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

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

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

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

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

               Now substitute $u$ back in:

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

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

         1. The integral of cosine is sine:

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

         The result is: $- \frac{\sin^{7}{\left(x \right)}}{7} + \frac{3 \sin^{5}{\left(x \right)}}{5} - \sin^{3}{\left(x \right)} + \sin{\left(x \right)}$

      So, the result is: $\frac{144 \sin^{7}{\left(x \right)}}{7} - \frac{432 \sin^{5}{\left(x \right)}}{5} + 144 \sin^{3}{\left(x \right)} - 144 \sin{\left(x \right)}$

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

      $$\int 116 \cos^{5}{\left(x \right)}\, dx = 116 \int \cos^{5}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

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

      2. Rewrite the integrand:

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

      3. Integrate term-by-term:

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

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

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

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

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

            Now substitute $u$ back in:

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

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

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

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

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

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

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

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

               Now substitute $u$ back in:

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

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

         1. The integral of cosine is sine:

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

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

      So, the result is: $\frac{116 \sin^{5}{\left(x \right)}}{5} - \frac{232 \sin^{3}{\left(x \right)}}{3} + 116 \sin{\left(x \right)}$

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

      $$\int \left(- 40 \cos^{3}{\left(x \right)}\right)\, dx = - 40 \int \cos^{3}{\left(x \right)}\, dx$$

      1. Rewrite the integrand:

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

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

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

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

         1. Integrate term-by-term:

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

               $$\int 1\, du = u$$

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

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

               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: $- \frac{u^{3}}{3}$

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

         Now substitute $u$ back in:

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

      So, the result is: $\frac{40 \sin^{3}{\left(x \right)}}{3} - 40 \sin{\left(x \right)}$

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

      $$\int 5 \cos{\left(x \right)}\, dx = 5 \int \cos{\left(x \right)}\, dx$$

      1. The integral of cosine is sine:

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

      So, the result is: $5 \sin{\left(x \right)}$

   The result is: $\frac{64 \sin^{9}{\left(x \right)}}{9} - 16 \sin^{7}{\left(x \right)} + \frac{68 \sin^{5}{\left(x \right)}}{5} - \frac{16 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$

3. Now simplify:

   $$\frac{\left(320 \sin^{8}{\left(x \right)} - 720 \sin^{6}{\left(x \right)} + 612 \sin^{4}{\left(x \right)} - 240 \sin^{2}{\left(x \right)} + 45\right) \sin{\left(x \right)}}{45}$$

4. Add the constant of integration:

   $$\frac{\left(320 \sin^{8}{\left(x \right)} - 720 \sin^{6}{\left(x \right)} + 612 \sin^{4}{\left(x \right)} - 240 \sin^{2}{\left(x \right)} + 45\right) \sin{\left(x \right)}}{45}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(320 \sin^{8}{\left(x \right)} - 720 \sin^{6}{\left(x \right)} + 612 \sin^{4}{\left(x \right)} - 240 \sin^{2}{\left(x \right)} + 45\right) \sin{\left(x \right)}}{45}+ \mathrm{constant}$$