Daily math problems for June 9, 2026 — Trigonometry, Calculus

$$I = \int_0^{\pi/4} \ln(1 + \tan\theta)\,d\theta.$$

Reflect by $\phi = \pi/4 - \theta$. The angle-difference formula gives $\tan(\pi/4 - \phi) = \dfrac{1 - \tan\phi}{1 + \tan\phi}$, so

$$1 + \tan\theta \;=\; 1 + \frac{1 - \tan\phi}{1 + \tan\phi} \;=\; \frac{2}{1 + \tan\phi}.$$

Taking logs: $\;\ln(1 + \tan\theta) = \ln 2 - \ln(1 + \tan\phi)$.

Pair the integral with its reflected twin. Renaming the dummy variable $\phi \to \theta$ (the bounds are unchanged because the substitution is a reflection of $[0, \pi/4]$ onto itself):

$$I \;=\; \int_0^{\pi/4} \bigl[\ln 2 - \ln(1 + \tan\theta)\bigr]\,d\theta \;=\; \frac{\pi}{4}\ln 2 \;-\; I.$$

Solve. $\;2I = \dfrac{\pi}{4}\ln 2$, so

$$\boxed{\,I = \frac{\pi}{8}\ln 2\,.}$$

The reflection trick — pairing $f(\theta)$ with $f(\pi/4 - \theta)$ — works whenever the integrand simplifies under that reflection. Worth keeping in your toolbox alongside the half-angle and Weierstrass substitutions.