MHTCET 15 May 2024 Shift 2 Paper

Concept:Simplify the integrand by algebraic manipulation and then use the symmetry property of definite integrals.Explanation:Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx$.Multiply the numerator and denominator by $\sqrt{\sin x \cos x}$.The numerator becomes $\sqrt{\cot x} \cdot \sqrt{\sin x \cos x} = \cos x$.The denominator becomes $\cos x + \sin x$.Therefore, $I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \, dx$.Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$, replace $x$ by $\frac{\pi}{2} - x$.This gives $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} \, dx$.Adding the two expressions for $I$: $2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx = \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}$.Hence, $I = \frac{\pi}{4}$.Answer:$\frac{\pi}{4}$, which corresponds to option A.