Integral Calculus Question 270
Question: $ \int_{{}}^{{}}{\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx=} $
[RPET 1995]
Options:
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 $
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Answer:
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}{ {{\cot }^{-1}}({{\tan }^{2}}x) }=-\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}{ {{\tan }^{-1}}({{\tan }^{2}}x) }=\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x} $ .
BETA
