MCQ
If ${I_n} = \int_0^{\pi /4} {{{\tan }^n}\theta \,d\theta ,} $ then ${I_8} + {I_6}$ equals
- A$\frac{1}{4}$
- B$\frac{1}{5}$
- C$\frac{1}{6}$
- ✓$\frac{1}{7}$
${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 ${I_8} + {I_6} = \frac{1}{{8 - 1}} = \frac{1}{7}$.
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