$\displaystyle\int_0^{\pi/4} \tan^4 x\, dx = ?$
- $\dfrac{\pi}{4} - \dfrac{2}{3}$
- $\dfrac{\pi}{4}$
- $\dfrac{1}{3}$
- $\dfrac{1}{3} - \dfrac{\pi}{4} + \dfrac{\pi}{2}$... — actually $\dfrac{3\pi - 8}{12}$ form
Show answer and solution
Correct answer: A. $\dfrac{\pi}{4} - \dfrac{2}{3}$
**Reduction:** $I_n = \int_0^{\pi/4} \tan^n x\, dx$, $I_n + I_{n-2} = 1/(n-1)$. $I_0 = \pi/4$. $I_2 = 1 - \pi/4$.
$I_4 = 1/3 - I_2 = 1/3 - 1 + \pi/4 = \pi/4 - 2/3$.