`int_(0)^(pi//2)log(tanx+cotx)dx=pi(log2)`

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \log(\tan x + \cot x) \, dx \) and show that it equals \( \pi \log 2 \), we can follow these steps:

### Step 1: Rewrite the integrand
We start with the expression inside the logarithm:
\[ \tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \]
Since \(\sin^2 x + \cos^2 x = 1\), we can simplify:
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