If u_n = _0^ /4 ^n x ,dx, then u_n + u_ n - 2 = - Vidyadip

MCQ

If ${u_n} = \int_0^{\pi /4} {{{\tan }^n}x\,dx,} $ then ${u_n} + {u_{n - 2}} = $

✓
$\frac{1}{{n - 1}}$
B
$\frac{1}{{n + 1}}$
C
$\frac{1}{{2n - 1}}$
D
$\frac{1}{{2n + 1}}$

✓

Answer

Correct option: A.

$\frac{1}{{n - 1}}$

a
(a) ${u_n} = \int_0^{\pi /4} {{{\tan }^n}x\,dx} $

$ = \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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