NDA 1 2024 Maths, question 78

Solution.

Answer: (a) $\frac{\pi}{4}$

1. Replacing $x$ by $\frac{\pi}{2} - x$ swaps $\sin x$ and $\cos x$: $I = \int_{0}^{\pi/2}\frac{a + \cos x}{2a + \sin x + \cos x}\,dx$.
2. Adding the two forms: $2I = \int_{0}^{\pi/2}\frac{2a + \sin x + \cos x}{2a + \sin x + \cos x}\,dx = \frac{\pi}{2}$.
3. So $I = \frac{\pi}{4}$.

Trap: This needs the denominator never to be zero on $[0, \frac{\pi}{2}]$, true for example when $a \geq 0$.