MCQ
$\int_{}^{} {\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}} dx = $
  • $2[\sin x + x\cos \alpha ] + c$
  • B
    $2[\sin x + \sin \alpha ] + c$
  • C
    $2[ - \sin x + x\cos \alpha ] + c$
  • D
    $ - 2[\sin x + \sin \alpha ] + c$

Answer

Correct option: A.
$2[\sin x + x\cos \alpha ] + c$
a
(a)$\int_{}^{} {\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}} \,dx = \int_{}^{} {\frac{{2({{\cos }^2}x - {{\cos }^2}\alpha )}}{{\cos x - \cos \alpha }}\,dx} $
$ = 2\int_{}^{} {(\cos x + \cos \alpha )\,dx} = 2(\sin x + x\cos \alpha )$.

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