Question

$\int_{0}^{\pi/2} \log\left( \frac{\cos x}{\sin x} \right)\, dx$ is equal to:

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\pi$
• D. $2\pi$
• E. $0$

Hint:

For definite integrals involving ratios of $\sin$ and $\cos$ within a logarithm from $0$ to $\pi/2$, the result is almost always zero because the "Kings Property" flips the ratio, and their sum results in $\log(1)$.

Answer 1

Answer: (E)

Concept: To solve definite integrals of the form $\int_{0}^{a} f(x) \, dx$, we can use the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$.

Step 1: Apply the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$.
Let $I = \int_{0}^{\pi/2} \log\left( \frac{\cos x}{\sin x} \right) \, dx$.
Applying the property where $a = \pi/2$: \[ I = \int_{0}^{\pi/2} \log\left( \frac{\cos(\pi/2 - x)}{\sin(\pi/2 - x)} \right) \, dx \] \[ I = \int_{0}^{\pi/2} \log\left( \frac{\sin x}{\cos x} \right) \, dx \]

Step 2: Add the two expressions for $I$.
\[ 2I = \int_{0}^{\pi/2} \left[ \log\left( \frac{\cos x}{\sin x} \right) + \log\left( \frac{\sin x}{\cos x} \right) \right] \, dx \] Using the property $\log A + \log B = \log(AB)$: \[ 2I = \int_{0}^{\pi/2} \log\left( \frac{\cos x}{\sin x} \cdot \frac{\sin x}{\cos x} \right) \, dx \] \[ 2I = \int_{0}^{\pi/2} \log(1) \, dx \]

Step 3: Conclusion.
Since $\log 1 = 0$: \[ 2I = \int_{0}^{\pi/2} 0 \, dx = 0 \Rightarrow I = 0 \]