A) \[{{\cot }^{-1}}({{\tan }^{2}}x)+c\]
B) \[{{\tan }^{-1}}({{\tan }^{2}}x)+c\]
C) \[{{\cot }^{-1}}({{\cot }^{2}}x)+c\]
D) \[{{\tan }^{-1}}({{\cot }^{2}}x)+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}\,dx}\] \[=\int_{{}}^{{}}{\frac{2\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}\,dx=\int_{{}}^{{}}{\frac{2\tan x{{\sec }^{2}}x}{1+{{\tan }^{4}}x}\,dx}}\] Put \[{{\tan }^{2}}x=t\Rightarrow 2\tan x{{\sec }^{2}}x\,dx=dt,\] then it reduced to \[\Rightarrow f(x)=\frac{{{x}^{3}}}{3}+5x+c\]. 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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