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: Use the reflection property.
For a definite integral from $0$ to $\frac{\pi}{2}$, replacing $x$ with $\frac{\pi}{2} - x$ does not change the value. This swaps $\tan x$ into $\cot x$, which is the key here.

Step 2: Name the integral.
Let the integral be $I$.
\[ I = \int_0^{\pi/2}\ln(\tan x)\,dx \]

Step 3: Apply the property.
Replace $x$ with $\frac{\pi}{2} - x$. Since $\tan\left(\frac{\pi}{2} - x\right) = \cot x$, we get a new form.
\[ I = \int_0^{\pi/2}\ln(\cot x)\,dx \]

Step 4: Add the two forms.
Add the original $I$ and this new $I$. Logs add to give a product inside.
\[ 2I = \int_0^{\pi/2}\left[\ln(\tan x) + \ln(\cot x)\right]dx = \int_0^{\pi/2}\ln(\tan x\cdot\cot x)\,dx \]

Step 5: Simplify the product.
Since $\tan x$ and $\cot x$ are reciprocals, their product is $1$, and $\ln 1 = 0$.
\[ 2I = \int_0^{\pi/2} 0\,dx = 0 \]

Step 6: Solve for $I$.
So $2I = 0$, which gives $I = 0$.
\[ \boxed{0} \]