int_0^{pi /2} {frac{{dx}}{{2 + cos x}}} = | vedclass

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(C) Let $I = \int_0^{\pi /2} {\frac{{dx}}{{2 + \cos x}}} $. Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \sin^2 \fra

Vedclass · Accepted Answer

$\frac{2}{{\sqrt 3 }}{	an ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} ight)$

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(C) Let $I = \int_0^{\pi /2} {\frac{{dx}}{{2 + \cos x}}} $. Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$,we have $2 + \cos x = 2(\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}) + \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = \sin^2 \frac{x}{2} + 3\cos^2 \frac{x}{2}$. Thus,$I = \int_0^{\pi /2} {\frac{{dx}}{{\sin^2 \frac{x}{2} + 3\cos^2 \frac{x}{2}}}} $. Dividing numerator and denominator by $\cos^2 \frac{x}{2}$,we get $I = \int_0^{\pi /2} {\frac{{\sec^2 \frac{x}{2}}}{{3 + 	an^2 \frac{x}{2}}}} dx$. Let $t = 	an \frac{x}{2}$,then $dt = \frac{1}{2} \sec^2 \frac{x}{2} dx$,which implies $\sec^2 \frac{x}{2} dx = 2 dt$. When $x = 0$,$t = 0$. When $x = \pi/2$,$t = 	an(\pi/4) = 1$. Substituting these,$I = \int_0^1 \frac{2 dt}{3 + t^2} = 2 \int_0^1 \frac{dt}{(\sqrt{3})^2 + t^2}$. Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a})$,we get $I = 2 [\frac{1}{\sqrt{3}} 	an^{-1}(\frac{t}{\sqrt{3}})]_0^1 = \frac{2}{\sqrt{3}} 	an^{-1}(\frac{1}{\sqrt{3}})$. Therefore,the correct option is $C$.

$\int_0^{\pi /2} {\frac{{dx}}{{2 + \cos x}}} = $

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