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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
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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