Integral of sin^3xcosxdx dx

1. There are multiple ways to do this integral.

   Method #1

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

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

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

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

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

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

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

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

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

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

      1. Integrate term-by-term:

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

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

            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: $\frac{u^{2}}{4}$

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

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

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

      Now substitute $u$ back in:

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

2. Add the constant of integration:

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

The answer is: