Biết ( int limits_0^{ frac{x}{4}} {{{ tan }^2}x ;dx} = a - frac{ pi }{b} left( {a,b ; in Z} right). )Tính (S = a + {b^2}. ) Đáp án: ___ 17

( int limits_0^{ frac{x}{4}} {{{ tan }^2}x ;dx} = int limits_0^{ frac{x}{4}} { left( {{{ tan }^2}x + 1 - 1} right)dx} = int limits_0^{ frac{ pi }{4}} { frac{1}{{{{ cos }^2}x}} ;dx} - int limits_0^{ frac{ pi }{4}} {1 ;dx} = 1 - frac{ pi }{4} ) ( Rightarrow left { { begin{array}{*{20}{l}}{a = 1} {b = 4} end{array}} right. Rightarrow a + {b^2} = 17. )