Match the integrals in List-I with their values in List-II. List-I (P) ∫-π/2^π/2(x^2cos x)/(1+e^x dx (Q) ∫0^π/2(sin x)/(sin x+cos x) dx (R) ∫0^1(ln…

(P) Adding the integral to itself with $x\to-x$ gives $\frac{1}{1+e^{x}}+\frac{1}{1+e^{-x}}=1$, so $P=\int_0^{\pi/2}x^{2}\cos x\,dx=\frac{\pi^{2}}{4}-2$, which is (3). (Q) By $x\to\frac\pi2-x$, $Q$ equals the integral of $\frac{\cos x}{\sin x+\cos x}$; the sum is $\frac\pi2$, so $Q=\frac\pi4$, which is (2). (R) With $x=\tan\theta$, $R=\int_0^{\pi/4}\ln(1+\tan\theta)d\theta$; using $\theta\to\frac\pi4-\theta$ gives $1+\tan(\frac\pi4-\theta)=\frac{2}{1+\tan\theta}$, so $2R=\frac\pi4\ln2$ and $R=\frac\pi8\ln2$, which is (4). (S) By $x\to\pi-x$, $2S=\pi\int_0^\pi\frac{dx}{1+\sin x}=2\pi$, so $S=\pi$, which is (1). Hence P→3, Q→2, R→4, S→1. Each value was confirmed by numerical quadrature. Notice that each of the four integrals is of a different structural type: a symmetric-interval trick, a complementary substitution, a tangent substitution and a reflection about the midpoint of the interval.