UPSC NDA 2 2022 Mathematics Solved Paper

Concept:This integral simplifies by splitting the numerator into two separate fractions and then using standard substitution methods.Explanation:We start with the given integral:$I = \int \frac{(\cos x)^{1.5} - (\sin x)^{1.5}}{\sqrt{\sin x \cos x}} \, dx$Rewrite each term separately:$I = \int \left( \frac{\cos x}{\sqrt{\sin x}} - \frac{\sin x}{\sqrt{\cos x}} \right) dx$This becomes $I = I_1 - I_2$, where$I_1 = \int \frac{\cos x}{\sqrt{\sin x}} \, dx$ and $I_2 = \int \frac{\sin x}{\sqrt{\cos x}} \, dx$For $I_1$, let $\sin x = t$. Then $\cos x \, dx = dt$.So $I_1 = \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} \, dt = 2\sqrt{t} + c_1 = 2\sqrt{\sin x} + c_1$For $I_2$, let $\cos x = u$. Then $-\sin x \, dx = du \implies \sin x \, dx = -du$.So $I_2 = \int \frac{-du}{\sqrt{u}} = -\int u^{-1/2} \, du = -2\sqrt{u} + c_2 = -2\sqrt{\cos x} + c_2$Now subtract:$I = I_1 - I_2 = (2\sqrt{\sin x} + c_1) - (-2\sqrt{\cos x} + c_2)$$= 2\sqrt{\sin x} + 2\sqrt{\cos x} + (c_1 - c_2)$Combine constants into a single constant $c$:$I = 2\sqrt{\sin x} + 2\sqrt{\cos x} + c$Answer:$\int\frac{(\cos x)^{1.5}-(\sin x)^{1.5}}{\sqrt{\sin x\cdot\cos x}}dx = 2\sqrt{\sin x} + 2\sqrt{\cos x} + c$Thus, the correct option is C.