The value of $$\int_0^{20\pi} (\sin^4 x + \cos^4 x)\, dx$$ is equal to :

Write the integrand in a simpler form.
Using $$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x\cos^2 x$$ and $$\sin^2 x + \cos^2 x = 1$$, we get

$$\sin^4 x + \cos^4 x = 1 - 2\sin^2 x\cos^2 x$$

Next, use $$\sin^2 x\cos^2 x = \left(\tfrac12\sin 2x\right)^2 = \tfrac14\sin^2 2x$$.
Therefore

$$\sin^4 x + \cos^4 x = 1 - \tfrac12\sin^2 2x$$

The required integral becomes

$$I = \int_{0}^{20\pi} \left[1 - \tfrac12\sin^2 2x\right]\,dx = \int_{0}^{20\pi} 1\,dx \;-\; \tfrac12 \int_{0}^{20\pi} \sin^2 2x\,dx$$

1. Integral of the constant: $$\int_{0}^{20\pi} 1\,dx = 20\pi$$.

2. Evaluate $$\int_{0}^{20\pi} \sin^2 2x\,dx$$.
Write $$\sin^2 2x = \tfrac12 (1 - \cos 4x)$$, so

$$\int_{0}^{20\pi} \sin^2 2x\,dx = \tfrac12\int_{0}^{20\pi} (1 - \cos 4x)\,dx = \tfrac12\left[ x - \tfrac{\sin 4x}{4} \right]_{0}^{20\pi}$$

Since $$\sin 4x$$ is periodic of period $$\tfrac{\pi}{2}$$, $$\sin 4(20\pi)=\sin(80\pi)=0$$ and $$\sin 0 = 0$$. Hence

$$\int_{0}^{20\pi} \sin^2 2x\,dx = \tfrac12 (20\pi - 0) = 10\pi$$

Now put this back:

$$I = 20\pi - \tfrac12 (10\pi) = 20\pi - 5\pi = 15\pi$$

Thus the value of the integral is $$15\pi$$.

Option C which is: $$15\pi$$