dx 7 + 5 x = - Vidyadip

MCQ

$\int {\frac{{dx}}{{7 + 5\cos x}} = } $

✓
$\frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}\tan \frac{x}{2}} \right) + c$
B
$\frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}\tan \frac{x}{2}} \right) + c$
C
$\frac{1}{4}{\tan ^{ - 1}}\left( {\tan \frac{x}{2}} \right) + c$
D
$\frac{1}{7}{\tan ^{ - 1}}\left( {\tan \frac{x}{2}} \right) + c$

✓

Answer

Correct option: A.

$\frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}\tan \frac{x}{2}} \right) + c$

(a)$I = \frac{{dx}}{{7 + 5\cos x}}$$ = \int {\frac{{dx}}{{7 + 5\,\left( {\frac{{1 - {{\tan }^2}(x/2)}}{{1 + {{\tan }^2}(x/2)}}} \right)}}} $

$ = \int {\frac{{{{\sec }^2}(x/2)\,dx}}{{7 + 7{{\tan }^2}(x/2) + 5 - 5{{\tan }^2}(x/2)}}} $

$ = \int {\frac{{{{\sec }^2}(x/2)\,dx}}{{12 + 2{{\tan }^2}(x/2)}}} $$ = \int {\frac{{\frac{1}{2}{{\sec }^2}(x/2)\,.dx}}{{6 + {{\tan }^2}(x/2)}}} $

$\tan \frac{x}{2} = t$ $⇒$ $\frac{1}{2}{\sec ^2}\frac{x}{2}dx = dt$

$I = \int {\frac{{dt}}{{{t^2} + ({{\sqrt {6)} }^2}}}} $ $ = \frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\frac{t}{{\sqrt 6 }} + c$

$ = \frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\left| {\frac{{\tan (x/2)}}{{\sqrt 6 }}} \right| + c$

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