Question

The value of the definite integral $\int_0^{\pi/2} \ln(\tan x) \, dx$ is:

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

Hint:

Since the tangent and cotangent functions are reciprocals, their product inside a logarithm sums to $\ln(1) = 0$, giving an instant answer of $0$.

Answer 1

The Correct Option is A

Step 1: Concept
We use the definite integral property $\int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx$.

Step 2: Meaning
For the limits $0$ to $\frac{\pi}{2}$, we replace $x$ with $\frac{\pi}{2} - x$. This converts $\tan x$ into $\cot x$.

Step 3: Analysis
Let the given integral be $I$: \[ I = \int_0^{\pi/2} \ln(\tan x) \, dx \quad \text{--- (Eq. 1)} \] Applying the property: \[ I = \int_0^{\pi/2} \ln\left(\tan\left(\frac{\pi}{2} - x\right)\right) \, dx = \int_0^{\pi/2} \ln(\cot x) \, dx \quad \text{--- (Eq. 2)} \] Adding Equation 1 and Equation 2: \[ 2I = \int_0^{\pi/2} \left[ \ln(\tan x) + \ln(\cot x) \right] \, dx \] \[ 2I = \int_0^{\pi/2} \ln(\tan x \cdot \cot x) \, dx \] Since $\tan x \cdot \cot x = 1$ and $\ln(1) = 0$: \[ 2I = \int_0^{\pi/2} \ln(1) \, dx = \int_0^{\pi/2} 0 \, dx = 0 \implies I = 0 \]

Step 4: Conclusion
The value of the definite integral is $0$. Final Answer: (A)