Question

Evaluate : $\underset{0}{\overset{π/2}{∫}}\frac{1}{3+2cosx}dx$

• A. $\sqrt{5}ta{n}^{−1}\left(\frac{1}{\sqrt{5}}\right)$
• B. $\frac{\sqrt{5}}{2}ta{n}^{−1}\left(\frac{1}{\sqrt{5}}\right)$
• C. $\frac{2}{\sqrt{5}}ta{n}^{−1}\left(\frac{1}{\sqrt{5}}\right)$
• D. $−\frac{2}{\sqrt{5}}ta{n}^{−1}\left(\frac{1}{\sqrt{5}}\right)$

Answer 1

Answer: (C)

$I=\underset{0}{\overset{π/2}{∫}}\frac{1}{3+2cosx}dx$
$=\underset{0}{\overset{π/2}{∫}}\frac{se{c}^2\frac{x}{2}}{5+ta{n}^2\frac{x}{2}}dx$
Let $tan\frac{x}{2}=t⇒\frac{1}{2}se{c}^2\frac{x}{2}dx=dt$
Now, $x=0⇒t=0$and $x=\frac{π}{2}⇒t=1$
$∴I=\underset{0}{\overset{1}{∫}}\frac{2dt}{5+t^2}=2â‹\dots \frac{1}{\sqrt{5}}{\left[ta{n}^{−1}\left(\frac{t}{\sqrt{5}}\right)\right]}_0^1$
$=\frac{2}{\sqrt{5}}\left[ta{n}^{−1}\frac{1}{\sqrt{5}}−ta{n}^{−1}0\right]=\frac{2}{\sqrt{5}}ta{n}^{−1}\left(\frac{1}{\sqrt{5}}\right)$