Evaluate: ∫₀^π (sin⁴ x + cos⁴ x) dx Hence evaluate: ∫_(-2π)^(2π) (sin⁴ x + cos⁴ x)/(1 + eˣ) dx

$\int_0^\pi (\sin^4 x + \cos^4 x)\,dx$ $\left[\because \sin^4 x + \cos^4 x = 1 - \frac{1}{2}\sin^2 2x = \frac{1}{4}(3 + \cos 4x)\right]$ $= \frac{1}{4}\int_0^\pi (3 + \cos 4x)\,dx = \frac{1}{4}\left[3x + \frac{\sin 4x}{4}\right]_0^\pi = \frac{3\pi}{4}$ $I = \int_{-2\pi}^{2\pi} \frac{\sin^4 x + \cos^4 x}{1 + e^x}\,dx = \int_{-2\pi}^{2\pi} \frac{\sin^4(-x) + \cos^4(-x)}{1 + e^{-x}}\,dx$ $\left[\because \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx\right]$ $= \int_{-2\pi}^{2\pi} \frac{\sin^4 x + \cos^4 x}{1 + e^{-x}}\,dx = \int_{-2\pi}^{2\pi} \frac{e^x(\sin^4 x + \cos^4 x)}{1 + e^x}\,dx$ $I + I = \int_{-2\pi}^{2\pi} \frac{\sin^4 x + \cos^4 x}{1 + e^x}\,dx + \int_{-2\pi}^{2\pi} \frac{e^x(\sin^4 x + \cos^4 x)}{1 + e^x}\,dx$ $\therefore 2I = \int_{-2\pi}^{2\pi}(\sin^4 x + \cos^4 x)\,dx = 2\int_0^{2\pi}(\sin^4 x + \cos^4 x)\,dx$ $\left[\because \int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx \text{ if } f(-x) = f(x)\right]$ $\therefore I = \int_0^{2\pi}(\sin^4 x + \cos^4 x)\,dx = 2\int_0^\pi(\sin^4 x + \cos^4 x)\,dx$ $\left[\because \int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx \text{ if } f(2a-x) = f(x)\right]$ $= 2 \cdot \frac{3\pi}{4} = \frac{3\pi}{2}$