Question

Evaluate the integral $\displaystyle \int_0^{\frac{\pi{2}} \log(\tan x + \cot x)\, dx$}

• A. $\pi \log 2$
• B. $-\pi \log 2$
• C. $\dfrac{\pi}{2} \log 2$
• D. $2\pi \log 2$

Hint:

Use symmetry in definite integrals and properties like \( \int_0^a f(x)dx = \int_0^a f(a-x)dx \) to simplify complex logarithmic integrals.

Answer 1

The Correct Option is A

We use the property of definite integrals: \[ \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx \] Let \( I = \int_0^{\frac{\pi}{2}} \log(\tan x + \cot x)\,dx \) Then using the property: \[ I = \int_0^{\frac{\pi}{2}} \log(\tan(\frac{\pi}{2} - x) + \cot(\frac{\pi}{2} - x))\, dx = \int_0^{\frac{\pi}{2}} \log(\cot x + \tan x)\, dx = I \] So adding both, \[ 2I = \int_0^{\frac{\pi}{2}} \left[\log(\tan x + \cot x) + \log(\cot x + \tan x)\right]\, dx = 2 \int_0^{\frac{\pi}{2}} \log(\tan x + \cot x)\, dx \] Let’s substitute \( \tan x + \cot x = \dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} = \dfrac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \dfrac{1}{\sin x \cos x} \) So, \[ I = \int_0^{\frac{\pi}{2}} \log\left(\frac{1}{\sin x \cos x}\right)\, dx = -\int_0^{\frac{\pi}{2}} \log(\sin x \cos x)\, dx \] \[ = -\int_0^{\frac{\pi}{2}} \log\left(\frac{1}{2} \sin 2x\right)\, dx = -\int_0^{\frac{\pi}{2}} \log\left(\frac{1}{2}\right)\, dx - \int_0^{\frac{\pi}{2}} \log(\sin 2x)\, dx \] \[ = \left(\frac{\pi}{2} \log 2 \right) - \frac{1}{2} \int_0^{\pi} \log(\sin u)\, du = \left(\frac{\pi}{2} \log 2\right) - \frac{1}{2}(-\pi \log 2) \] \[ = \left(\frac{\pi}{2} \log 2\right) + \left(\frac{\pi}{2} \log 2\right) = \pi \log 2 \]