<p>Which is correct for \(I_n=\int_0^{\pi/2}(\sin x+\cos x)^n\,dx\)? [JEE Advanced 2013]</p>
Step-by-Step Solution
Key Concept: I_2 = \int_0^(\pi/2)(sinx+cosx)^2dx = \int_0^(\pi/2)(1+sin2x)dx = \pi/2+[-cos2x/2]_0^(\pi/2) = \pi/2+0 = \pi/2. I_1=\int_0^(\pi/2)\sqrt{2} \cdot sin(x+\pi/4)dx=\sqrt{2} \cdot [-cos(x+\pi/4)]_0^(\pi/2)=\sqrt{2} \cdot (1/\sqrt{2}+1/\sqrt{2})=2.
Step 1: Evaluate $I_0$.
$$I_0 = \int_0^{\pi/2} (\sin x + \cos x)^0 dx = \int_0^{\pi/2} 1 \, dx = [x]_0^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$
Step 2: Evaluate $I_1$.
$$I_1 = \int_0^{\pi/2} (\sin x + \cos x)^1 dx = [-\cos x + \sin x]_0^{\pi/2} = (-\cos(\pi/2) + \sin(\pi/2)) - (-\cos(0) + \sin(0)) = (0 + 1) - (-1 + 0) = 1 - (-1) = 2$$
Step 3: Evaluate $I_2$.
$$I_2 = \int_0^{\pi/2} (\sin x + \cos x)^2 dx = \int_0^{\pi/2} (\sin^2 x + \cos^2 x + 2 \sin x \cos x) dx = \int_0^{\pi/2} (1 + \sin 2x) dx$$
$$I_2 = [x - \frac{\cos 2x}{2}]_0^{\pi/2} = \left(\frac{\pi}{2} - \frac{\cos \pi}{2}\right) - \left(0 - \frac{\cos 0}{2}\right)$$
$$I_2 = \left(\frac{\pi}{2} - \frac{-1}{2}\right) - \left(0 - \frac{1}{2}\right) = \frac{\pi}{2} + \frac{1}{2} + \frac{1}{2} = \frac{\pi}{2} + 1$$
Correct Answer: B