91Ó°ÊÓ

We will use the power-reduction formula to rewrite \(\cos^2(8\theta)\) in terms of \(\cos\) and sine functions of a single angle. The power reduction formula for \(\cos^2{x}\) is: $$\cos^2{x} = \frac{1 + \cos{2x}}{2}$$ Applying the formula to our function: $$\cos^2(8\theta) = \frac{1 + \cos{16\theta}}{2}$$ So, the integral becomes: $$\int_{0}^{\pi / 4} \frac{1 + \cos{16\theta}}{2} d\theta$$

We will integrate the function with respect to \(\theta\). Find the integral of each term inside the brackets: $$\int \frac{1}{2} d\theta = \frac{1}{2}\theta + C_1$$ $$\int \frac{\cos{16\theta}}{2} d\theta = \frac{\sin{16\theta}}{32} + C_2$$ So, the integral of the function is: $$\frac{1}{2}\theta + \frac{\sin{16\theta}}{32} + C$$

Finally, we will apply the limits of integration from the given interval to find the definite integral. The interval is given as \([0, \frac{\pi}{4}]\), so we will substitute these values into the integrand: $$\left[\frac{1}{2}\theta + \frac{\sin{16\theta}}{32}\right]_{0}^{\pi / 4}$$ Evaluate the integral at the upper limit: $$\frac{1}{2} \cdot \frac{\pi}{4} + \frac{\sin{16 \cdot \frac{\pi}{4}}}{32} = \frac{\pi}{8}$$ Evaluate the integral at the lower limit: $$\frac{1}{2} \cdot 0 + \frac{\sin{16 \cdot 0}}{32} = 0$$ Subtract the lower limit evaluation from the upper limit evaluation: $$\frac{\pi}{8} - 0 = \frac{\pi}{8}$$ So, the definite integral of the function \(\cos^2(8\theta)\) over the interval \([0, \frac{\pi}{4}]\) is: $$\int_{0}^{\pi / 4} \cos ^{2} 8 \theta d \theta = \frac{\pi}{8}$$