MCQ
જો ${u_n} = \int_0^{\pi /4} {{{\tan }^n}x\,dx,} $ તો ${u_n} + {u_{n - 2}} = $
- ✓$\frac{1}{{n - 1}}$
- B$\frac{1}{{n + 1}}$
- C$\frac{1}{{2n - 1}}$
- D$\frac{1}{{2n + 1}}$
$ = \int_0^{\pi /4} {({{\sec }^2}x - 1){{\tan }^{n - 2}}x\,\,dx} $
$ = \int_0^{\pi /4} {{{\sec }^2}x{{\tan }^{n - 2}}x\,\,dx} - \int_0^{\pi /4} {{{\tan }^{n - 2}}x\,\,dx} $
$ = \left[ {\frac{{{{\tan }^{n - 1}}x}}{{n - 1}}} \right]_0^{\pi /4} - {u_{n - 2}}$
$ \Rightarrow {u_n} + {u_{n - 2}} = \frac{1}{{n - 1}}$.
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