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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{1}{{1 + {{\sin }^2}x}}\;dx = } $
  • ✓
    $\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
  • B
    $\sqrt 2 {\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
  • C
    $ - \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
  • D
    $ - \sqrt 2 {\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
✓

Answer

Correct option: A.
$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
a
(a) $I = \int_{}^{} {\frac{1}{{1 + {{\sin }^2}x}}\,dx} = \int_{}^{} {\frac{{dx}}{{2{{\sin }^2}x + {{\cos }^2}x}}} $
$ = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{2{{\tan }^2}x + 1}}} $$ = \frac{1}{2}\int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{{{\tan }^2}x + \frac{1}{2}}}} $
Put $\tan x = t \Rightarrow {\sec ^2}x\,dx = dt,$ then
$I = \frac{1}{2}\int_{}^{} {\frac{{dt}}{{{t^2} + \frac{1}{2}}} = \frac{1}{2}} .\frac{1}{{1\sqrt 2 }}{\tan ^{ - 1}}\frac{t}{{1\sqrt 2 }}$
$ = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$.

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