Question

Evaluate : $\int\limits_{0}^{\pi /2} \frac{1}{3+2\,cos \,x}dx$

• A. $\sqrt{5} \,tan^{-1}\left(\frac{1}{\sqrt{5}}\right)$
• B. $\frac{\sqrt{5}}{2} \,tan^{-1}\left(\frac{1}{\sqrt{5}}\right)$
• C. $\frac{2}{\sqrt{5}} \,tan^{-1}\left(\frac{1}{\sqrt{5}}\right)$
• D. $-\frac{2}{\sqrt{5}} \,tan^{-1}\left(\frac{1}{\sqrt{5}}\right)$

Answer 1

Answer: (C)

$I= \int\limits_{0}^{\pi /2} \frac{1}{3+2\, cos\, x} dx $
$= \int\limits_{0}^{\pi /2} \frac{sec^{2} \frac{x}{2} }{5+tan^{2} \frac{x}{2}} dx $
Let $tan \frac{x}{2} = t \Rightarrow \frac{1}{2} sec^{2} \frac{x}{2} dx = dt $
Now, $x=0 \Rightarrow t =0 \,\,$and $\,\,x=\frac{\pi}{2} \Rightarrow t=1 $
$ \therefore I =\int\limits_{0}^{1} \frac{2dt}{5+t^{2}} = 2\cdot\frac{1}{\sqrt{5}}\left[tan^{-1}\left(\frac{t}{\sqrt{5}}\right)\right]_{0}^{1}$
$= \frac{2}{\sqrt{5}} \left[tan^{-1} \frac{1}{\sqrt{5}} -tan ^{-1} 0\right] = \frac{2}{\sqrt{5}} tan^{-1}\left(\frac{1}{\sqrt{5}}\right)$