Question

${\int }_0^{\pi /2}\frac{1}{1+4{\sin }^{2}x}dx=$

• A. $\frac{\pi }{\sqrt{5}}$
• B. $\frac{\pi }{2\sqrt{5}}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{3\sqrt{5}}$

Answer 1

Answer: (B)

$\frac{\pi }{2\sqrt{5}}$

${\int }_0^{\frac{\pi }{4}}\frac{1}{1+4si{n}^{2}x}dx={\int }_0^{\frac{\pi }{4}}\frac{1}{1+4co{s}^{2}x}dx$
${\int }_0^{\frac{\pi }{4}}\frac{1}{1+4co{s}^{2}x}dx={\int }_0^{\frac{\pi }{4}}\frac{se{c}^{2}x}{(se{c}^{2}x+4)}dx$
$={\int }_0^{\frac{\pi }{4}}\frac{d\left(tanx\right)}{5+ta{n}^{2}x}dx={[\frac{1}{\sqrt{5}}ta{n}^{-1}\left(\frac{tanx}{\sqrt{5}}\right)+c]}_0^{\frac{\pi }{2}}$
$=\frac{1}{\sqrt{5}}\left(\frac{\pi }{2}\right)$
$=\frac{\pi }{2\sqrt{5}}$