Evaluate Definite Integral of Sin to Power 4 x Plus Cos to Power 4 x

To evaluate the definite integral

$I = \int_{0}^{20\pi} (\sin^4 x + \cos^4 x) \, \mathrm{d}x$

we can first simplify the integrand using standard trigonometric identities.

Step 1: Simplify the Integrand

Using the algebraic identity $a^2 + b^2 = (a+b)^2 - 2ab$, where $a = \sin^2 x$ and $b = \cos^2 x$:

$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$

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

$\sin^4 x + \cos^4 x = 1 - 2\left(\frac{\sin(2x)}{2}\right)^2 = 1 - \frac{1}{2}\sin^2(2x)$

Now, applying the half-angle formula $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$ with $\theta = 2x$:

$\sin^2(2x) = \frac{1 - \cos(4x)}{2}$

Substitute this back into the expression:

$\sin^4 x + \cos^4 x = 1 - \frac{1}{2} \left( \frac{1 - \cos(4x)}{2} \right) = 1 - \frac{1}{4} + \frac{1}{4}\cos(4x) = \frac{3}{4} + \frac{1}{4}\cos(4x)$

Step 2: Evaluate the Integral

Substitute the simplified expression back into the definite integral:

$I = \int_{0}^{20\pi} \left( \frac{3}{4} + \frac{1}{4}\cos(4x) \right) \mathrm{d}x$

Integrating term-by-term:

$I = \left[ \frac{3}{4}x + \frac{1}{16}\sin(4x) \right]_{0}^{20\pi}$

Now, substitute the upper and lower limits:

$I = \left( \frac{3}{4}(20\pi) + \frac{1}{16}\sin(80\pi) \right) - \left( \frac{3}{4}(0) + \frac{1}{16}\sin(0) \right)$

Since $\sin(80\pi) = 0$ and $\sin(0) = 0$:

$I = 15\pi + 0 - 0 = 15\pi$

Conclusion

The value of the definite integral is $15\pi$.

Hence, the correct option is C.