The value of $\displaystyle\int_0^{\pi/2} \log\!\left(\dfrac{4+3\sin x}{4+3\cos x}\right)dx$ is
A$2$
B$\dfrac{3}{4}$
C$0$
D$-2$
Answer & Solution
Correct answer: C. $0$
1. Let $I$ be the integral. Apply $\mathbf{P_4}$ ($x\to\tfrac{\pi}{2}-x$): $\sin$ and $\cos$ swap.
2. So $I=\int_0^{\pi/2}\log\!\left(\dfrac{4+3\cos x}{4+3\sin x}\right)dx=-I$.
3. Hence $2I=0$, giving $I=0$.
4. The swap turns the log argument into its reciprocal, producing $-I$.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.347_
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