7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

${I_n} = \int_0^{\pi /4} {{{\sec }^2}\theta {{\tan }^{n - 2}}\theta \,d\theta } - \int_0^{\pi /2} {{{\tan }^{n - 2}}\theta } \,d\theta $

${I_n} = \left[ {\frac{{{{\tan }^{n - 1}}\theta }}{{n - 1}}} \right]_0^{\pi /4} - {I_{n - 2}} $

$\Rightarrow {I_n} + {I_{n - 2}} = \frac{1}{{n - 1}}$

Hence ${I_8} + {I_6} = \frac{1}{{8 - 1}} = \frac{1}{7}$.