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

Integrate : $\int \large\frac{\cos 2x - \cos 2 \alpha}{\cos x - \cos \alpha}$$ dx $

\[\begin {array} {1 1} (a)\;2 (\sin x + x \sin \alpha)+c \\ (b)\; 2 ( \sin \alpha + x \cos \alpha)+c \\ (c)\;2 ( \sin x + x \cos \alpha)+c \\ (d)\;2 (\sin x + \sin \alpha )+c \end {array}\]

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1 Answer

By adding and subtracting (1) in numerator
$\qquad= \int \large\frac{(\cos 2x+1)-(\cos 2 \alpha +1)}{\cos x - \cos \alpha}.$$ dx$
$\qquad= 2 \int \large\frac{\cos^2 x - \cos^2 \alpha }{\cos x - \cos \alpha}$$.dx$
$\qquad = 2 \int \cos x + \cos \alpha .dx$
$\qquad= 2 \{ \sin x + x \cos \alpha \}+c$
Hence c is the correct answer.
answered Dec 21, 2013 by meena.p
 
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