Question

${\int }_0^{\frac{\pi }{2}}\frac{dx}{2+cosx}=$

• A. $\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• B. $\sqrt{3}ta{n}^{-1}\left(\sqrt{3}\right)$
• C. $\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• D. $2\sqrt{3}ta{n}^{-1}\left(\sqrt{3}\right)$

Answer 1

The correct option is C $\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{2+cosx}$
$={\int }_0^{\frac{\pi }{2}}\frac{dx}{2si{n}^2\frac{x}{2}+2co{s}^2\frac{x}{2}+co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}}$
$={\int }_0^{\frac{\pi }{2}}\frac{dx}{si{n}^2\frac{x}{2}+3co{s}^2\frac{x}{2}}={\int }_0^{\frac{\pi }{2}}\frac{se{c}^2\frac{x}{2}}{3+ta{n}^2\frac{x}{2}}dx$
$Putt=tan\frac{x}{2}\Rightarrow dt=\frac{1}{2}se{c}^2\frac{x}{2}dx,then$
$I=2{\int }_0^1\frac{dt}{3+t^2}=\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$