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

(Note: This question has been split into 2 questions) This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

1 Answer

Toolbox:
  • $\cos 2x = cos^2x - 1$
Given $\int\large \frac{cos 2x-cos 2\alpha}{cos x-cos \alpha}$$dx$
We know that $\cos 2x = cos^2x - 1$
$\Rightarrow \int\large \frac{cos 2x-cos 2\alpha}{cos x-cos \alpha}\normalsize dx=\int\large\frac{(cos^2x-1)-(2cos^2\alpha-1)}{cos x-cos\alpha}$
$\Rightarrow \int\large \frac{cos 2x-cos 2\alpha}{cos x-cos \alpha}\normalsize dx=\int \large \frac{2(cos^2x-cos^2\alpha)}{cos x-cos\alpha}$dx.
$\Rightarrow \int\large \frac{cos 2x-cos 2\alpha}{cos x-cos \alpha}\normalsize dx$$=2\int (cos x+cos\alpha)$$dx =2\int cos x dx+2\int cos\alpha$$dx$..
$\Rightarrow \int\large \frac{cos 2x-cos 2\alpha}{cos x-cos \alpha}\normalsize dx$$=2sin x+2xcos\alpha$ +c.
answered Mar 21, 2013 by sreemathi.v
edited Mar 22, 2013 by balaji.thirumalai
 

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