UPSC NDA 2 2016 Mathematics Solved Paper

Concept:The integral is solved by splitting it into simpler parts and applying integration by parts together with substitution.Explanation:Step 1: Write the given integral as$I = \int e^{\sin x}\left( \frac{x\cos^3 x - \sin x}{\cos^2 x} \right)dx$Separate the fraction:$\frac{x\cos^3 x - \sin x}{\cos^2 x} = x\cos x - \sec x \tan x$Thus, $I = \int e^{\sin x} x\cos x \, dx - \int e^{\sin x} \sec x \tan x \, dx$.Denote $I_1 = \int e^{\sin x} x\cos x \, dx$ and $I_2 = \int e^{\sin x} \sec x \tan x \, dx$.Step 2: Evaluate $I_1$ using integration by parts with $u = x$ and $dv = e^{\sin x}\cos x \, dx$.First, find $\int e^{\sin x}\cos x \, dx$ using substitution $t = \sin x$, $dt = \cos x\,dx$, giving $\int e^t dt = e^{\sin x}$.Now by parts: $I_1 = x e^{\sin x} - \int (1) e^{\sin x} dx = x e^{\sin x} - \int e^{\sin x} dx$.Step 3: Evaluate $I_2$ using integration by parts with $u = e^{\sin x}$ and $dv = \sec x \tan x \, dx$.We know $\int \sec x \tan x \, dx = \sec x$.Also $\frac{d}{dx}(e^{\sin x}) = e^{\sin x}\cos x$.So by parts: $I_2 = e^{\sin x} \sec x - \int (e^{\sin x}\cos x)(\sec x) dx = e^{\sin x}\sec x - \int e^{\sin x} dx$.Step 4: Substitute $I_1$ and $I_2$ back into $I$:$I = \left( x e^{\sin x} - \int e^{\sin x} dx \right) - \left( e^{\sin x}\sec x - \int e^{\sin x} dx \right)$The indefinite integrals $\int e^{\sin x} dx$ cancel, leaving $I = x e^{\sin x} - e^{\sin x}\sec x + c$.Factor $e^{\sin x}$: $I = (x - \sec x)e^{\sin x} + c$.Answer:The integral equals $(x - \sec x)e^{\sin x} + c$, which corresponds to option B.