Definite-Integration Question 512
Question: $ \int_0^{\pi /2}{{{\sin }^{2m}}x,dx=} $
Options:
A) $ \frac{2m!}{{{(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
Show Answer
Answer:
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} $ .
BETA
