Integral of sin(7*x)*cos(4*x) dx

Rewrite the integrand:

Integrate term-by-term:

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

   $$\int \left(- 512 \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 512 \int \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

   1. Rewrite the integrand:

      $$\sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{3} \sin{\left(x \right)} \cos^{4}{\left(x \right)}$$

   2. There are multiple ways to do this integral.

      Method #1

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

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

         $$\int \left(u^{10} - 3 u^{8} + 3 u^{6} - u^{4}\right)\, du$$

         1. Integrate term-by-term:

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

               $$\int u^{10}\, du = \frac{u^{11}}{11}$$

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

               $$\int \left(- 3 u^{8}\right)\, du = - 3 \int u^{8}\, du$$

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

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

               So, the result is: $- \frac{u^{9}}{3}$

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

               $$\int 3 u^{6}\, du = 3 \int u^{6}\, du$$

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

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

               So, the result is: $\frac{3 u^{7}}{7}$

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

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

               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: $- \frac{u^{5}}{5}$

            The result is: $\frac{u^{11}}{11} - \frac{u^{9}}{3} + \frac{3 u^{7}}{7} - \frac{u^{5}}{5}$

         Now substitute $u$ back in:

         $$\frac{\cos^{11}{\left(x \right)}}{11} - \frac{\cos^{9}{\left(x \right)}}{3} + \frac{3 \cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$$

      Method #2

      1. Rewrite the integrand:

         $$\left(1 - \cos^{2}{\left(x \right)}\right)^{3} \sin{\left(x \right)} \cos^{4}{\left(x \right)} = - \sin{\left(x \right)} \cos^{10}{\left(x \right)} + 3 \sin{\left(x \right)} \cos^{8}{\left(x \right)} - 3 \sin{\left(x \right)} \cos^{6}{\left(x \right)} + \sin{\left(x \right)} \cos^{4}{\left(x \right)}$$

      2. Integrate term-by-term:

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

            $$\int \left(- \sin{\left(x \right)} \cos^{10}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{10}{\left(x \right)}\, dx$$

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

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

               $$\int u^{10}\, du$$

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

                  $$\int \left(- u^{10}\right)\, du = - \int u^{10}\, du$$

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

                     $$\int u^{10}\, du = \frac{u^{11}}{11}$$

                  So, the result is: $- \frac{u^{11}}{11}$

               Now substitute $u$ back in:

               $$- \frac{\cos^{11}{\left(x \right)}}{11}$$

            So, the result is: $\frac{\cos^{11}{\left(x \right)}}{11}$

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

            $$\int 3 \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx = 3 \int \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx$$

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

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

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

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

                  $$\int \left(- u^{8}\right)\, du = - \int u^{8}\, du$$

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

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

                  So, the result is: $- \frac{u^{9}}{9}$

               Now substitute $u$ back in:

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

            So, the result is: $- \frac{\cos^{9}{\left(x \right)}}{3}$

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

            $$\int \left(- 3 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - 3 \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$$

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

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

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

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

                  $$\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$$

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

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

                  So, the result is: $- \frac{u^{7}}{7}$

               Now substitute $u$ back in:

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

            So, the result is: $\frac{3 \cos^{7}{\left(x \right)}}{7}$

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

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

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

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

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

               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:

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

         The result is: $\frac{\cos^{11}{\left(x \right)}}{11} - \frac{\cos^{9}{\left(x \right)}}{3} + \frac{3 \cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$

      Method #3

      1. Rewrite the integrand:

         $$\left(1 - \cos^{2}{\left(x \right)}\right)^{3} \sin{\left(x \right)} \cos^{4}{\left(x \right)} = - \sin{\left(x \right)} \cos^{10}{\left(x \right)} + 3 \sin{\left(x \right)} \cos^{8}{\left(x \right)} - 3 \sin{\left(x \right)} \cos^{6}{\left(x \right)} + \sin{\left(x \right)} \cos^{4}{\left(x \right)}$$

      2. Integrate term-by-term:

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

            $$\int \left(- \sin{\left(x \right)} \cos^{10}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{10}{\left(x \right)}\, dx$$

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

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

               $$\int u^{10}\, du$$

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

                  $$\int \left(- u^{10}\right)\, du = - \int u^{10}\, du$$

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

                     $$\int u^{10}\, du = \frac{u^{11}}{11}$$

                  So, the result is: $- \frac{u^{11}}{11}$

               Now substitute $u$ back in:

               $$- \frac{\cos^{11}{\left(x \right)}}{11}$$

            So, the result is: $\frac{\cos^{11}{\left(x \right)}}{11}$

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

            $$\int 3 \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx = 3 \int \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx$$

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

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

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

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

                  $$\int \left(- u^{8}\right)\, du = - \int u^{8}\, du$$

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

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

                  So, the result is: $- \frac{u^{9}}{9}$

               Now substitute $u$ back in:

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

            So, the result is: $- \frac{\cos^{9}{\left(x \right)}}{3}$

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

            $$\int \left(- 3 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - 3 \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$$

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

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

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

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

                  $$\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$$

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

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

                  So, the result is: $- \frac{u^{7}}{7}$

               Now substitute $u$ back in:

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

            So, the result is: $\frac{3 \cos^{7}{\left(x \right)}}{7}$

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

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

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

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

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

               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:

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

         The result is: $\frac{\cos^{11}{\left(x \right)}}{11} - \frac{\cos^{9}{\left(x \right)}}{3} + \frac{3 \cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$

   So, the result is: $- \frac{512 \cos^{11}{\left(x \right)}}{11} + \frac{512 \cos^{9}{\left(x \right)}}{3} - \frac{1536 \cos^{7}{\left(x \right)}}{7} + \frac{512 \cos^{5}{\left(x \right)}}{5}$

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

   $$\int 512 \sin^{7}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 512 \int \sin^{7}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$

   1. Rewrite the integrand:

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

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

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

      $$\int \left(u^{8} - 3 u^{6} + 3 u^{4} - u^{2}\right)\, du$$

      1. Integrate term-by-term:

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

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

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

            $$\int \left(- 3 u^{6}\right)\, du = - 3 \int u^{6}\, du$$

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

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

            So, the result is: $- \frac{3 u^{7}}{7}$

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

            $$\int 3 u^{4}\, du = 3 \int u^{4}\, du$$

            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: $\frac{3 u^{5}}{5}$

         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: $\frac{u^{9}}{9} - \frac{3 u^{7}}{7} + \frac{3 u^{5}}{5} - \frac{u^{3}}{3}$

      Now substitute $u$ back in:

      $$\frac{\cos^{9}{\left(x \right)}}{9} - \frac{3 \cos^{7}{\left(x \right)}}{7} + \frac{3 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$$

   So, the result is: $\frac{512 \cos^{9}{\left(x \right)}}{9} - \frac{1536 \cos^{7}{\left(x \right)}}{7} + \frac{1536 \cos^{5}{\left(x \right)}}{5} - \frac{512 \cos^{3}{\left(x \right)}}{3}$

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

   $$\int \left(- 64 \sin^{7}{\left(x \right)}\right)\, dx = - 64 \int \sin^{7}{\left(x \right)}\, dx$$

   1. Rewrite the integrand:

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

   2. Rewrite the integrand:

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

   3. Integrate term-by-term:

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

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

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

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

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

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

               $$\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$$

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

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

               So, the result is: $- \frac{u^{7}}{7}$

            Now substitute $u$ back in:

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

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

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

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

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

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

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

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

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

               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:

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

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

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

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

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

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

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

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

            Now substitute $u$ back in:

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

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

      1. The integral of sine is negative cosine:

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

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

   So, the result is: $- \frac{64 \cos^{7}{\left(x \right)}}{7} + \frac{192 \cos^{5}{\left(x \right)}}{5} - 64 \cos^{3}{\left(x \right)} + 64 \cos{\left(x \right)}$

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

   $$\int 896 \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 896 \int \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

   1. Rewrite the integrand:

      $$\sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} \cos^{4}{\left(x \right)}$$

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

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

      $$\int \left(- u^{8} + 2 u^{6} - u^{4}\right)\, du$$

      1. Integrate term-by-term:

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

            $$\int \left(- u^{8}\right)\, du = - \int u^{8}\, du$$

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

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

            So, the result is: $- \frac{u^{9}}{9}$

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

            $$\int 2 u^{6}\, du = 2 \int u^{6}\, du$$

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

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

            So, the result is: $\frac{2 u^{7}}{7}$

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

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

            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: $- \frac{u^{5}}{5}$

         The result is: $- \frac{u^{9}}{9} + \frac{2 u^{7}}{7} - \frac{u^{5}}{5}$

      Now substitute $u$ back in:

      $$- \frac{\cos^{9}{\left(x \right)}}{9} + \frac{2 \cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$$

   So, the result is: $- \frac{896 \cos^{9}{\left(x \right)}}{9} + 256 \cos^{7}{\left(x \right)} - \frac{896 \cos^{5}{\left(x \right)}}{5}$

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

   $$\int \left(- 896 \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 896 \int \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$

   1. Rewrite the integrand:

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

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

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

      $$\int \left(- u^{6} + 2 u^{4} - u^{2}\right)\, du$$

      1. Integrate term-by-term:

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

            $$\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$$

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

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

            So, the result is: $- \frac{u^{7}}{7}$

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

            $$\int 2 u^{4}\, du = 2 \int u^{4}\, du$$

            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: $\frac{2 u^{5}}{5}$

         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: $- \frac{u^{7}}{7} + \frac{2 u^{5}}{5} - \frac{u^{3}}{3}$

      Now substitute $u$ back in:

      $$- \frac{\cos^{7}{\left(x \right)}}{7} + \frac{2 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$$

   So, the result is: $128 \cos^{7}{\left(x \right)} - \frac{1792 \cos^{5}{\left(x \right)}}{5} + \frac{896 \cos^{3}{\left(x \right)}}{3}$

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

   $$\int 112 \sin^{5}{\left(x \right)}\, dx = 112 \int \sin^{5}{\left(x \right)}\, dx$$

   1. Rewrite the integrand:

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

   2. Rewrite the integrand:

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

   3. Integrate term-by-term:

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

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

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

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

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

            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 $- du$:

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

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

            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: $- \frac{112 \cos^{5}{\left(x \right)}}{5} + \frac{224 \cos^{3}{\left(x \right)}}{3} - 112 \cos{\left(x \right)}$

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

   $$\int \left(- 448 \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 448 \int \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

   1. Rewrite the integrand:

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

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

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

      $$\int \left(u^{6} - u^{4}\right)\, du$$

      1. Integrate term-by-term:

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

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

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

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

            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: $- \frac{u^{5}}{5}$

         The result is: $\frac{u^{7}}{7} - \frac{u^{5}}{5}$

      Now substitute $u$ back in:

      $$\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$$

   So, the result is: $- 64 \cos^{7}{\left(x \right)} + \frac{448 \cos^{5}{\left(x \right)}}{5}$