MHTCET 15 Apr 2026 Shift 1 Paper

Concept:Use the half-angle identity to simplify $\sqrt{1+\sin x}$ and then integrate using the standard sine integral form.Explanation:We need to evaluate the integral $\int \sqrt{2} \sqrt{1+\sin x} \, dx$ and match it with the given form $-4 \cos(ax+b)+c$. Use the known identity $1+\sin x = \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2$. Taking the positive square root branch, we get $\sqrt{1+\sin x} = \sin \frac{x}{2} + \cos \frac{x}{2}$. Now convert this sum into a single sine using the angle addition formula: $\sin \frac{x}{2} + \cos \frac{x}{2} = \sqrt{2} \sin \left(\frac{x}{2} + \frac{\pi}{4}\right)$. Substituting back into the integrand gives $\sqrt{2} \cdot \sqrt{2} \sin \left(\frac{x}{2} + \frac{\pi}{4}\right) = 2 \sin \left(\frac{x}{2} + \frac{\pi}{4}\right)$. Now integrate using the standard formula $\int 2 \sin(px+q) \, dx = -\frac{2}{p} \cos(px+q) + c$. Here $p = \frac{1}{2}$, so $-\frac{2}{p} = -4$. Therefore, $\int 2 \sin \left(\frac{x}{2} + \frac{\pi}{4}\right) dx = -4 \cos \left(\frac{x}{2} + \frac{\pi}{4}\right) + c$. Since the RHS has the structure $-4 \cos(ax+b)+c$, compare the angle $\frac{x}{2}+\frac{\pi}{4}$ with $ax+b$. So, $\frac{x}{2}$ matches $ax$ giving $a = \frac{1}{2}$, and the constant $\frac{\pi}{4}$ matches $b$. Option A has the correct $a$ but wrong $b$, while option C is invalid since it contains $x$.Answer:$a = \frac{1}{2}$ and $b = \frac{\pi}{4}$. Therefore, the correct option is B: $\left(\frac{1}{2}, \frac{\pi}{4}\right)$.