_0^ /2 ^ 2m x ,dx = - Vidyadip

MCQ

$\int_0^{\pi /2} {{{\sin }^{2m}}x\,dx = } $

A
$\frac{{2\,\,m\,\,!}}{{{{({2^m}.\,m\,\,!)}^2}}}.\frac{\pi }{2}$
✓
$\frac{{(2m)\,\,!}}{{{{({2^m}.\,m\,\,!)}^2}}}.\frac{\pi }{2}$
C
$\frac{{2m\,\,!}}{{{2^m}.\,{{(m\,\,!)}^2}}}.\frac{\pi }{2}$
D
None of these

✓

Answer

Correct option: B.

$\frac{{(2m)\,\,!}}{{{{({2^m}.\,m\,\,!)}^2}}}.\frac{\pi }{2}$

b
(b) Here the power is even, so from formula

$\int_0^{\pi /2} {{{\sin }^{2m}}} xdx = \frac{{(2m - 1)}}{{2m}}.\frac{{(2m - 3)}}{{(2m - 2)}}.....\frac{3}{4}.\frac{1}{2}.\frac{\pi }{2}$

$ = \frac{{2m.(2m - 1)(2m - 2)....3.2.1.\frac{\pi }{2}}}{{{{[2m.(2m - 2)(2m - 4).....4.2]}^2}}}$

Multiplying the numerator and the denominator by $2m(2m - 2)....4.2$

$ = \frac{{(2m)!}}{{{{[{2^m}.m(m - 1)(m - 2).....2.1]}^2}}}\frac{\pi }{2}$

$ = \frac{{(2m)!}}{{{{({2^m}.m!)}^2}}}\frac{\pi }{2}$ .

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