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

MCQ

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

A
${\cot ^{ - 1}}({\tan ^2}x) + c$
✓
${\tan ^{ - 1}}({\tan ^2}x) + c$
C
${\cot ^{ - 1}}({\cot ^2}x) + c$
D
${\tan ^{ - 1}}({\cot ^2}x) + c$

✓

Answer

Correct option: B.

${\tan ^{ - 1}}({\tan ^2}x) + c$

b
(b)$\int_{}^{} {\frac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}\,dx} $
$ = \int_{}^{} {\frac{{2\sin x\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}\,dx = \int_{}^{} {\frac{{2\tan x{{\sec }^2}x}}{{1 + {{\tan }^4}x}}\,dx} } $
Put ${\tan ^2}x = t \Rightarrow 2\tan x{\sec ^2}x\,dx = dt,$ then it reduced to $ \int {dt\over{1 + t^2}} = tan^{-1} t + c = tan^{-1}(tan^2 x) + c$.
Trick : By inspection,
$\frac{d}{{dx}}\left\{ {{{\cot }^{ - 1}}({{\tan }^2}x)} \right\} = - \frac{{1(2\tan x\,.\,{{\sec }^2}x)}}{{1 + {{\tan }^4}x}} = - \frac{{\sin 2x}}{{{{\cos }^4}x + {{\sin }^4}x}}$
$ \Rightarrow \frac{d}{{dx}}\left\{ {{{\tan }^{ - 1}}({{\tan }^2}x)} \right\} = \frac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}$.

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