UPSC NDA I 2018 Mathematics Solved Paper

Concept:Integration by parts and the ILATE rule are used to solve integrals involving product of exponential and trigonometric functions.Explanation:Let $I = \int_{0}^{\pi} e^{x} \sin x \, dx$.Apply integration by parts: let $u = \sin x$, $dv = e^{x}dx$.Then $du = \cos x \, dx$, $v = e^{x}$.So $I = \left[ \sin x \cdot e^{x} \right]_{0}^{\pi} - \int_{0}^{\pi} \cos x \cdot e^{x} \, dx$.At $x=0$, $\sin 0 = 0$; at $x=\pi$, $\sin\pi = 0$. Thus first term vanishes: $I = - \int_{0}^{\pi} \cos x \cdot e^{x} \, dx$.Now apply integration by parts again on $J = \int_{0}^{\pi} \cos x \, e^{x} dx$. Let $u = \cos x$, $dv = e^{x}dx$.Then $du = -\sin x \, dx$, $v = e^{x}$.So $J = \left[ \cos x \cdot e^{x} \right]_{0}^{\pi} - \int_{0}^{\pi} (-\sin x) e^{x} dx = \left[ \cos x \, e^{x} \right]_{0}^{\pi} + \int_{0}^{\pi} \sin x \, e^{x} dx = \left[ \cos x \, e^{x} \right]_{0}^{\pi} + I$.Therefore $I = - \left( \left[ \cos x \, e^{x} \right]_{0}^{\pi} + I \right)$. Substitute the limits: $\cos\pi = -1$, $\cos 0 = 1$, so $\left[ \cos x \, e^{x} \right]_{0}^{\pi} = (-1)e^{\pi} - (1)e^{0} = -e^{\pi} - 1$.Thus $I = - \left( -e^{\pi} - 1 + I \right) = e^{\pi} + 1 - I$.Bring $I$ terms together: $2I = e^{\pi} + 1$. Hence $I = \frac{e^{\pi} + 1}{2}$.Answer:$\frac{e^{\pi} + 1}{2}$ (Option A).