_ ^ dx 4 ^2 x + 5 ^2 x = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{4{{\sin }^2}x + 5{{\cos }^2}x}} = } $

A
$\frac{1}{{\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \right) + c$
B
$\frac{1}{{\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{\tan x}}{{\sqrt 5 }}} \right) + c$
✓
$\frac{1}{{2\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \right) + c$
D
None of these

✓

Answer

Correct option: C.

$\frac{1}{{2\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \right) + c$

c
(c)$\int_{}^{} {\frac{{dx}}{{4{{\sin }^2}x + 5{{\cos }^2}x}}} = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{4{{\tan }^2}x + 5}} = \frac{1}{4}\int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{{{\tan }^2}x + \frac{5}{4}}}} } $
Put $\tan x = t \Rightarrow {\sec ^2}x\,dx = dt,$ then it reduces to
$\frac{1}{4}\int_{}^{} {\frac{{dt}}{{{t^2} + {{\left( {\frac{{\sqrt 5 }}{2}} \right)}^2}}} = \frac{2}{{4\sqrt 5 }}{{\tan }^{ - 1}}\left( {\frac{{2t}}{{\sqrt 5 }}} \right)} + c$
$ = \frac{1}{{2\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \right) + c.$

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