A) \[\frac{2\,\,m\,\,!}{{{({{2}^{m}}.\,m\,\,!)}^{2}}}.\frac{\pi }{2}\]
B) \[\frac{(2m)\,\,!}{{{({{2}^{m}}.\,m\,\,!)}^{2}}}.\frac{\pi }{2}\]
C) \[\frac{2m\,\,!}{{{2}^{m}}.\,{{(m\,\,!)}^{2}}}.\frac{\pi }{2}\]
D) None of these
Correct Answer: B
Solution :
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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