Question:

Step 1: Calculate $I_0$. $$I_0 = \int_0^{\pi/2} (\sin x + \cos x)^0 dx = \int_0^{\pi/2} 1 \, dx = [x]_0^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$ Step 2: Calculate $I_1$. $$I_1 = \int_0^{\pi/2} (\sin x + \cos x) dx = [-\cos x + \sin x]_0^{\pi/2}$$ $$I_1 = (-\cos(\pi/2) + \sin(\pi/2)) - (-\cos(0) + \sin(0))$$ $$I_1 = (0 + 1) - (-1 + 0) = 1 - (-1) = 2$$ Step 3: Calculate $I_2$. $$I_2 = \int_0^{\pi/2} (\sin x + \cos x)^2 dx$$ $$I_2 = \int_0^{\pi/2} (\sin^2 x + \cos^2 x + 2\sin x \cos x) dx$$ Using the identities $\sin^2 x + \cos^2 x = 1$ and $2\sin x \cos x = \sin 2x$: $$I_2 = \int_0^{\pi/2} (1 + \sin 2x) dx$$ $$I_2 = [x - \frac{\cos 2x}{2}]_0^{\pi/2}$$ $$I_2 = \left(\frac{\pi}{2} - \frac{\cos(2 \cdot \pi/2)}{2}\right) - \left(0 - \frac{\cos(2 \cdot 0)}{2}\right)$$ $$I_2 = \left(\frac{\pi}{2} - \frac{\cos \pi}{2}\right) - \left(0 - \frac{\cos 0}{2}\right)$$ $$I_2 = \left(\frac{\pi}{2} - \frac{-1}{2}\right) - \left(-\frac{1}{2}\right)$$ $$I_2 = \frac{\pi}{2} + \frac{1}{2} + \frac{1}{2} = \frac{\pi}{2} + 1$$