Integral of (sinx)^7 dx

1. Rewrite the integrand:

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

2. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

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

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

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

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

      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 $- du$:

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

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

      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 $- du$:

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

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

      1. The integral of sine is negative cosine:

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

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

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

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

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

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

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

      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 $- du$:

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

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

      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 $- du$:

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

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

      1. The integral of sine is negative cosine:

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

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

3. Add the constant of integration:

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

The answer is: