`int_(pi)^((3pi)/2)(sin^(4)x+cos^(4)x)dx`

Answer

To solve the integral

\[ I = \int_{\pi}^{\frac{3\pi}{2}} \left(\sin^4 x + \cos^4 x\right) \, dx, \]

we can start by rewriting \(\sin^4 x + \cos^4 x\) using the identity for the sum of squares:

\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x. \]

Since \(\sin^2 x + \cos^2 x = 1\), we have:

\[ \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x. \]

Using the double angle identity, we can express \(\sin^2 x \cos^2 x\) as:

\[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x). \]

Thus, we can rewrite our integral as:

\[ I = \int_{\pi}^{\frac{3\pi}{2}} \left(1 - \frac{1}{2} \sin^2(2x)\right) \, dx. \]

Now, we can separate the integral:

\[ I = \int_{\pi}^{\frac{3\pi}{2}} 1 \, dx - \frac{1}{2} \int_{\pi}^{\frac{3\pi}{2}} \sin^2(2x) \, dx. \]

Calculating the first integral:

\[ \int_{\pi}^{\frac{3\pi}{2}} 1 \, dx = \left[x\right]_{\pi}^{\frac{3\pi}{2}} = \frac{3\pi}{2} - \pi = \frac{\pi}{2}. \]

Next, we need to compute the second integral, \(\int_{\pi}^{\frac{3\pi}{2}} \sin^2(2x) \, dx\). We can use the identity:

\[ \sin^2(2x) = \frac{1 - \cos(4x)}{2}. \]

Thus,

\[ \int_{\pi}^{\frac{3\pi}{2}} \sin^2(2x) \, dx = \int_{\pi}^{\frac{3\pi}{2}} \frac{1 - \cos(4x)}{2} \, dx = \frac{1}{2} \int_{\pi}^{\frac{3\pi}{2}} 1 \, dx - \frac{1}{2} \int_{\pi}^{\frac{3\pi}{2}} \cos(4x) \, dx. \]

Calculating the first part again:

\[ \frac{1}{2} \int_{\pi}^{\frac{3\pi}{2}} 1 \, dx = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}. \]

Now for the second part:

\[ \int_{\pi}^{\frac{3\pi}{2}} \cos(4x) \, dx = \left[\frac{\sin(4x)}{4}\right]_{\pi}^{\frac{3\pi}{2}} = \frac{\sin(6\pi)}{4} - \frac{\sin(4\pi)}{4} = 0 - 0 = 0. \]

Thus,

\[ \int_{\pi}^{\frac{3\pi}{2}} \sin^2(2x) \, dx = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

Now substituting back into our expression for \(I\):

\[ I = \frac{\pi}{2} - \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{2} - \frac{\pi}{8} = \frac{4\pi}{8} - \frac{\pi}{8} = \frac{3\pi}{8}. \]

Thus, the final answer is:

\[ \boxed{\frac{3\pi}{8}}. \]