Question

$\int_{0}^{\pi/2}{\sin^{2m}xdx =}$

• 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

Answer 1

Answer: (B)

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

Answer 2

Here the power is even, so from formula

$\int_{0}^{\pi/2}\sin^{2m}xdx = \frac{(2m - 1)}{2m}.\frac{(2m - 3)}{(2m - 2)}.....\frac{3}{4}.\frac{1}{2}.\frac{\pi}{2}$

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

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

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

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