Usually for odd powers of the sine function, we use a little trick that relies on trigonometric identities and substitution. For instance, $\int\sin^3(x)dx = \int\sin(x)\sin^2(x) = \int\sin(x)(1-\cos^2(x))$, then using the substitution $u=\cos(x)$, we get $\int -(1-u^2)du\vert_{u=\cos(x)} = \cos^3(x)/3 - \cos(x) + c$.

In Numerical Methods That (Usually) Work, Forman S. Acton included a short interlude on what not to compute. This section is intended as a way to think carefully about numerical computation with a computer based on analytic solutions and rather think whether good enough approximations can be used.

As an example, consider the expression $\int_0^{0.3} \sin^8\theta\cdot d\theta$. We have

$$ \begin{equation} \begin{aligned} \int_0^{0.3} \sin^8\theta\cdot d\theta &= \left[(-\frac{1}{8}\cos\theta)(\sin^4\theta + \frac{7}{6}\sin^2\theta + \frac{35}{24})(\sin^3\theta) + \frac{105}{384}(\theta - \sin 2\theta)\right]_0^{0.3}\cr &= (-0.119417)(0.007627 + 0.101887 + 1.458333)(0.0258085) + 0.004341\cr &= -0.0048320 + 0.0048341 = 0.0000021 \end{aligned} \end{equation} $$

Note that this formula allows to compute a small number based on the difference between two much larger numbers. However, using a crude approximation for $\sin\theta$,¹ we get

$$ \int_0^{0.3} \theta^8\cdot d\theta = \frac{1}{9}\left[\theta^9\right]_0^{0.3} = 0.00000219. $$

A second term in the series can be added if more precision is required. You may also like Approximating the Sine Function.