# Integral de $\sin^{3}{\left(x \right)}$

A calculadora encontrará a integral/antiderivada de $\sin^{3}{\left(x \right)}$, com as etapas mostradas.

Por favor, escreva sem nenhum diferencial como $dx$, $dy$ etc.
Deixe em branco para detecção automática.

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Encontre $\int \sin^{3}{\left(x \right)}\, dx$.

### Solução

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 $\alpha=x$:

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

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

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

The integral can be rewritten as

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

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 $u=\cos{\left(x \right)}$:

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

Therefore,

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

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