Integral of $- \sin^{3}{\left(\theta \right)}$

Find $\int \left(- \sin^{3}{\left(\theta \right)}\right)\, d\theta$.

Apply the constant multiple rule $\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$ with $c=-1$ and $f{\left(\theta \right)} = \sin^{3}{\left(\theta \right)}$:

$${\color{red}{\int{\left(- \sin^{3}{\left(\theta \right)}\right)d \theta}}} = {\color{red}{\left(- \int{\sin^{3}{\left(\theta \right)} d \theta}\right)}}$$

Strip out one sine and write everything else in terms of the cosine, using the formula $\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$ with $\alpha=\theta$:

$$- {\color{red}{\int{\sin^{3}{\left(\theta \right)} d \theta}}} = - {\color{red}{\int{\left(1 - \cos^{2}{\left(\theta \right)}\right) \sin{\left(\theta \right)} d \theta}}}$$

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

Then $du=\left(\cos{\left(\theta \right)}\right)^{\prime }d\theta = - \sin{\left(\theta \right)} d\theta$ (steps can be seen »), and we have that $\sin{\left(\theta \right)} d\theta = - du$.

The integral becomes

$$- {\color{red}{\int{\left(1 - \cos^{2}{\left(\theta \right)}\right) \sin{\left(\theta \right)} d \theta}}} = - {\color{red}{\int{\left(u^{2} - 1\right)d u}}}$$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$ with $c=-1$ and $f{\left(u \right)} = 1 - u^{2}$:

$$- {\color{red}{\int{\left(u^{2} - 1\right)d u}}} = - {\color{red}{\left(- \int{\left(1 - u^{2}\right)d u}\right)}}$$

Integrate term by term:

$${\color{red}{\int{\left(1 - u^{2}\right)d u}}} = {\color{red}{\left(\int{1 d u} - \int{u^{2} d u}\right)}}$$

Apply the constant rule $\int c\, du = c u$ with $c=1$:

$$- \int{u^{2} d u} + {\color{red}{\int{1 d u}}} = - \int{u^{2} d u} + {\color{red}{u}}$$

Apply the power rule $\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$ $\left(n \neq -1 \right)$ with $n=2$:

$$u - {\color{red}{\int{u^{2} d u}}}=u - {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=u - {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

Recall that $u=\cos{\left(\theta \right)}$:

$${\color{red}{u}} - \frac{{\color{red}{u}}^{3}}{3} = {\color{red}{\cos{\left(\theta \right)}}} - \frac{{\color{red}{\cos{\left(\theta \right)}}}^{3}}{3}$$

Therefore,

$$\int{\left(- \sin^{3}{\left(\theta \right)}\right)d \theta} = - \frac{\cos^{3}{\left(\theta \right)}}{3} + \cos{\left(\theta \right)}$$

Add the constant of integration:

$$\int{\left(- \sin^{3}{\left(\theta \right)}\right)d \theta} = - \frac{\cos^{3}{\left(\theta \right)}}{3} + \cos{\left(\theta \right)}+C$$

$\int \left(- \sin^{3}{\left(\theta \right)}\right)\, d\theta = \left(- \frac{\cos^{3}{\left(\theta \right)}}{3} + \cos{\left(\theta \right)}\right) + C$A