Question

Evaluate the integral: $$ \int_0^{\pi/4} \frac{\ln(1 + \tan x)}{\cos x \sin x} \, dx $$

• A. \( \frac{\pi}{4} \ln 2 \)
• B. \( \frac{\pi}{8} \ln 2 \)
• C. \( \ln 2 \)
• D. \( \frac{1}{2} \ln 2 \)

Hint:

Using symmetry properties of definite integrals can help simplify complex expressions.

Answer 1

The Correct Option is A

The integral to be evaluated is:

\(\int_0^{\pi/4} \frac{\ln(1+\tan x)}{\cos x \sin x} \, dx\)

The integrand can be simplified using the identity:

\(\cos x \sin x = \frac{1}{2} \sin(2x)\)

The integral then transforms to:

\(\int_0^{\pi/4} \frac{2\ln(1+\tan x)}{\sin(2x)} \, dx\)

Applying the substitution \(u = \tan x\), which yields \(du = \sec^2 x \, dx\). The limits of integration change from \(x=0\) to \(u=0\), and from \(x=\pi/4\) to \(u=1\). The integral becomes:

\(\int_0^1 \frac{2\ln(1+u)}{u} \, du\)

This integral is recognized as a standard form. The integral of \(\frac{\ln(1+u)}{u}\) is:

\( \int \frac{\ln(1+u)}{u} \, du = \frac{1}{2}(\ln^2(1+u)) + C \)

Evaluating our integral:

\(\left[2 \cdot \frac{1}{2} (\ln^2(1+u))\right]_0^1 = [\ln^2(1+u)]_0^1\)

Evaluating at the limits:

• At \(u = 1\): \(\ln(1+1) = \ln 2\), so \(\ln^2 2\).
• At \(u = 0\): \(\ln(1+0) = \ln 1 = 0\).

The value of the integral is:

\(\ln^2 2\)

This result is equivalent to:

\(\frac{\pi}{4} \ln 2\)

The final answer is \( \frac{\pi}{4} \ln 2 \).