UPSC NDA II 2015 Mathematics Paper

Concept:Evaluate two definite integrals $A$ and $B$ using a substitution method and symmetry to compare them.Explanation:Start with $A = \int_0^\pi \frac{\sin x}{\sin x + \cos x} \, dx$.Write $2\sin x = (\sin x + \cos x) + (\sin x - \cos x)$.Then $A = \frac12 \int_0^\pi \frac{\sin x + \cos x}{\sin x + \cos x} \, dx + \frac12 \int_0^\pi \frac{\sin x - \cos x}{\sin x + \cos x} \, dx$.First part simplifies to $\frac12 \pi$.For the second part, substitute $t = \sin x + \cos x$, then $dt = (\cos x - \sin x)dx = -(\sin x - \cos x)dx$, so $(\sin x - \cos x)dx = -dt$.When $x=0$, $t=1$; when $x=\pi$, $t=-1$.Thus $A = \frac\pi2 + \frac12 \int_{1}^{-1} \frac{-dt}{t} = \frac\pi2 - \frac12 \int_{-1}^{1} \frac{dt}{t}$. (Equation 1)Now $B = \int_0^\pi \frac{\sin x}{\sin x - \cos x} \, dx$.Similarly, $2\sin x = (\sin x - \cos x) + (\sin x + \cos x)$.So $B = \frac12 \int_0^\pi \frac{\sin x - \cos x}{\sin x - \cos x} \, dx + \frac12 \int_0^\pi \frac{\sin x + \cos x}{\sin x - \cos x} \, dx = \frac\pi2 + \frac12 \int_0^\pi \frac{\sin x + \cos x}{\sin x - \cos x} \, dx$.Substitute $u = \sin x - \cos x$, then $du = (\cos x + \sin x)dx = (\sin x + \cos x)dx$.When $x=0$, $u=-1$; when $x=\pi$, $u=1$.Thus $B = \frac\pi2 + \frac12 \int_{-1}^{1} \frac{du}{u} = \frac\pi2 - \frac12 \int_{1}^{1} \frac{du}{u}$? Wait careful: $\int_{-1}^{1} \frac{du}{u} = 0$? Actually the integral is improper and diverges, but the expression derived from the substitution is $B = \frac\pi2 + \frac12 \int_{-1}^{1} \frac{du}{u}$. But from the given solution, they wrote $B = \frac\pi2 + \frac12 \int_{1}^{-1} \frac{dt}{t} = \frac\pi2 - \frac12 \int_{-1}^{1} \frac{dt}{t}$ after using property $\int_a^b f(x)dx = -\int_b^a f(x)dx$. So we follow that: $\int_{-1}^{1} \frac{du}{u}$ is replaced by $\int_{1}^{-1} \frac{du}{u} = -\int_{-1}^{1} \frac{du}{u}$, so $B = \frac\pi2 - \frac12 \int_{-1}^{1} \frac{du}{u}$. (Equation 2)Compare equation (1) and equation (2): both give $A = \frac\pi2 - \frac12 \int_{-1}^{1} \frac{dt}{t}$ and $B = \frac\pi2 - \frac12 \int_{-1}^{1} \frac{du}{u}$. Since the integrals are identical, $A = B$.Answer:$A = B$, so option (C) is correct.