Integral of sin(x)cos^3(x)dx dx

1. 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 $- du$:

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

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

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

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

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

         So, the result is: $- \frac{u^{4}}{4}$

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

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

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

      Then let $du = 2 \sin{\left(x \right)} \cos{\left(x \right)} dx$ and substitute $du$:

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

      1. Integrate term-by-term:

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

            $$\int \frac{1}{2}\, du = \frac{u}{2}$$

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

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

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

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

            So, the result is: $- \frac{u^{2}}{4}$

         The result is: $- \frac{u^{2}}{4} + \frac{u}{2}$

      Now substitute $u$ back in:

      $$- \frac{\sin^{4}{\left(x \right)}}{4} + \frac{\sin^{2}{\left(x \right)}}{2}$$

2. Add the constant of integration:

   $$- \frac{\cos^{4}{\left(x \right)}}{4}+ \mathrm{constant}$$

The answer is:

$$- \frac{\cos^{4}{\left(x \right)}}{4}+ \mathrm{constant}$$