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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $\large\frac{1+ \cos 4x }{\cot x - \tan x }$$dx$

\[\begin {array} {1 1} (a)\;-\frac{1}{8} \sin 4 x +c \\ (b)\;-\frac{1}{8} \cos 4x +c \\ (c)\;-\frac{1}{8}+ \cos 4x +c \\ (d)\;None \end {array}\]
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1 Answer

$\int \large\frac{2 \cos ^2 2x}{\cos ^2 x - \sin ^2 x} \times$$ \sin x \cos x$$ \;dx$
$\sin 2x = 2 \sin x \cos x$
$\cos 2x = \cos ^2 x - \sin ^2 x$
$\qquad= \int \cos 2x . \sin 2x .dx$
$\qquad=\large\frac{1}{2}$$ \int \sin 4x dx$
$\qquad=\large\frac{-1}{8}$$ \cos 4x +c$
Hence b is the correct answer.
answered Dec 21, 2013 by meena.p
 
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