Integral of sin(x)^9 dx

1. Rewrite the integrand:

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

2. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

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

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

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

            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:

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

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

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

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

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

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

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

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

               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: $\frac{4 \cos^{7}{\left(x \right)}}{7}$

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

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

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

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

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

            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:

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

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

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

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

         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: $\frac{4 \cos^{3}{\left(x \right)}}{3}$

      1. The integral of sine is negative cosine:

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

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

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

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

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

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

            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:

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

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

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

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

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

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

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

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

               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: $\frac{4 \cos^{7}{\left(x \right)}}{7}$

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

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

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

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

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

            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:

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

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

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

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

         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: $\frac{4 \cos^{3}{\left(x \right)}}{3}$

      1. The integral of sine is negative cosine:

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

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

3. Now simplify:

   $$\frac{\left(- 378 \sin^{4}{\left(x \right)} - 35 \cos^{8}{\left(x \right)} + 180 \cos^{6}{\left(x \right)} - 336 \cos^{2}{\left(x \right)} + 63\right) \cos{\left(x \right)}}{315}$$

4. Add the constant of integration:

   $$\frac{\left(- 378 \sin^{4}{\left(x \right)} - 35 \cos^{8}{\left(x \right)} + 180 \cos^{6}{\left(x \right)} - 336 \cos^{2}{\left(x \right)} + 63\right) \cos{\left(x \right)}}{315}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(- 378 \sin^{4}{\left(x \right)} - 35 \cos^{8}{\left(x \right)} + 180 \cos^{6}{\left(x \right)} - 336 \cos^{2}{\left(x \right)} + 63\right) \cos{\left(x \right)}}{315}+ \mathrm{constant}$$