_ ^ dx 5 + 4 x = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{5 + 4\cos x}} = } $

A
$\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{1}{3}\tan x} \right) + c$
B
$\frac{1}{3}{\tan ^{ - 1}}\left( {\frac{1}{3}\tan x} \right) + c$
✓
$\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{1}{3}\tan \frac{x}{2}} \right) + c$
D
$\frac{1}{3}{\tan ^{ - 1}}\left( {\frac{1}{3}\tan \frac{x}{2}} \right) + c$

✓

Answer

Correct option: C.

$\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{1}{3}\tan \frac{x}{2}} \right) + c$

c
(c)$\int_{}^{} {\frac{{dx}}{{5 + 4\cos x}}} $$ = \int_{}^{} {\frac{{dx}}{{5 + 4\left[ {\frac{{1 - {{\tan }^2}\frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}}} \right]}}} = \int_{}^{} {\frac{{{{\sec }^2}\frac{x}{2}}}{{9 + {{\tan }^2}\frac{x}{2}}}} \,dx$
Put $\tan \frac{x}{2} = t,$ then it reduces to
$2\int_{}^{} {\frac{{dt}}{{{3^2} + {t^2}}} = \frac{2}{3}{{\tan }^{ - 1}}\left[ {\frac{1}{3}\tan \frac{x}{2}} \right] + c} $
Aliter : Apply direct formula
i.e., $\int_{}^{} {\frac{1}{{a + b\cos x}}\,dx} $, {a > b}
$ = \frac{2}{{\sqrt {{a^2} - {b^2}} }}{\tan ^{ - 1}}\left[ {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{x}{2}} \right] + c$
We get $\int_{}^{} {\frac{{dx}}{{5 + 4\cos x}}} = \frac{2}{3}{\tan ^{ - 1}}\left\{ {\frac{1}{3}\tan \frac{x}{2}} \right\} + c.$

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