MathematicsMedium210×since 2002Q3324In=∫0π/4tannx dx{I_n} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx}In=0∫π/4tannxdx then limn→∞ n[In+In+2]\,\mathop {\lim }\limits_{n \to \infty } \,n\left[ {{I_n} + {I_{n + 2}}} \right]n→∞limn[In+In+2] equalsA12{1 \over 2}21B111C∞\infty∞DzeroCheck answerSkip