Linearize ( x)3 and calculate integral0^(pi)/(2) ( x)3 dx.

We know that $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$.
Therefore, $(\sin x)^3 = \left(\frac{e^{ix} - e^{-ix}}{2i}\right)^3$.
Expand the numerator using the binomial expansion $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

$$(e^{ix} - e^{-ix})^3 = (e^{ix})^3 - 3(e^{ix})^2(e^{-ix}) + 3(e^{ix})(e^{-ix})^2 - (e^{-ix})^3$$ $$= e^{i3x} - 3e^{i2x}e^{-ix} + 3e^{ix}e^{-i2x} - e^{-i3x}$$ $$= e^{i3x} - 3e^{ix} + 3e^{-ix} - e^{-i3x}$$

Group terms to form sine functions:

$$= (e^{i3x} - e^{-i3x}) - 3(e^{ix} - e^{-ix})$$

Using the identity $e^{i\theta} - e^{-i\theta} = 2i \sin \theta$:

$$= 2i \sin(3x) - 3(2i \sin x)$$

Now substitute this back into the expression for $(\sin x)^3$:

$$(\sin x)^3 = \frac{2i \sin(3x) - 6i \sin x}{(2i)^3}$$ $$= \frac{2i (\sin(3x) - 3 \sin x)}{-8i}$$ $$= \frac{\sin(3x) - 3 \sin x}{-4}$$ $$= -\frac{1}{4} \sin(3x) + \frac{3}{4} \sin x$$

So, the linearized form is:

$$(\sin x)^3 = \frac{3}{4} \sin x - \frac{1}{4} \sin(3x)$$

Substitute the linearized expression into the integral:

$$I = \int_0^{\frac{\pi}{2}} \left(\frac{3}{4} \sin x - \frac{1}{4} \sin(3x)\right) dx$$

Integrate term by term:

$$I = \frac{3}{4} \int_0^{\frac{\pi}{2}} \sin x \, dx - \frac{1}{4} \int_0^{\frac{\pi}{2}} \sin(3x) \, dx$$

The antiderivative of $\sin x$ is $-\cos x$.
The antiderivative of $\sin(3x)$ is $-\frac{1}{3}\cos(3x)$.

$$I = \left[-\frac{3}{4} \cos x - \frac{1}{4} \left(-\frac{1}{3} \cos(3x)\right)\right]_0^{\frac{\pi}{2}}$$ $$I = \left[-\frac{3}{4} \cos x + \frac{1}{12} \cos(3x)\right]_0^{\frac{\pi}{2}}$$

Evaluate the expression at the upper limit $x = \frac{\pi}{2}$:

$$\left(-\frac{3}{4} \cos\left(\frac{\pi}{2}\right) + \frac{1}{12} \cos\left(3 \cdot \frac{\pi}{2}\right)\right) = \left(-\frac{3}{4} \cdot 0 + \frac{1}{12} \cdot 0\right) = 0$$

Evaluate the expression at the lower limit $x = 0$:

$$\left(-\frac{3}{4} \cos(0) + \frac{1}{12} \cos(3 \cdot 0)\right) = \left(-\frac{3}{4} \cdot 1 + \frac{1}{12} \cdot 1\right) = -\frac{3}{4} + \frac{1}{12}$$

To combine these, find a common denominator, which is 12:

$$-\frac{3}{4} + \frac{1}{12} = -\frac{9}{12} + \frac{1}{12} = -\frac{8}{12} = -\frac{2}{3}$$

Subtract the lower limit value from the upper limit value:

$$I = 0 - \left(-\frac{2}{3}\right) = \frac{2}{3}$$

The final answer is $\mathbf{\frac{2}{3}}$.