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Home  >>  CBSE XII  >>  Math  >>  Integrals
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$\Large \int\frac{\cos 2x-\cos 2\theta}{\cos x-\cos\theta}\normalsize dx$ is equal to\begin{array}{1 1}(A)\;2(\sin x+x\cos\theta)+C & (B)\;2(\sin x-x\cos\theta)+C\\(C)\;2(\sin x+2x\cos\theta)+C & (D)\;2(\sin x-2x\cos\theta)+C\end{array}

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  • $\cos 2x=2\cos^2x-1$
  • $\int \cos x dx=\sin x+c$
Step 1:
$I=\Large \int\frac{\cos 2x-\cos 2\theta}{\cos x-\cos\theta}\normalsize dx$
$\cos 2x=2\cos^2x-1$
$I=\Large \int\frac{(2\cos ^2x-1)-(2\cos^2\theta-1)}{\cos x-\cos\theta}\normalsize dx$
On simplifying we get,
$\;\;\Large \int\frac{2(\cos ^2x-cos^2\theta)}{\cos x-\cos\theta}\normalsize dx$
$(\cos^2x-\cos^2\theta)=(\cos x+\cos \theta)(\cos x-\cos \theta)$
$\;\;=\int\large \frac{2(\cos x+\cos \theta)(\cos x-\cos\theta)}{\cos x-\cos \theta}\normalsize dx.$
$\;\;=2\int (\cos x+\cos\theta)dx.$
Step 2:
This can be splitted as
$I=2\begin{bmatrix}\int \cos x dx+\int \cos \theta.dx\end{bmatrix}$
On integrating we get,
$2(\sin x+x\cos\theta)+c$
Hence the correct option is (A)
answered Apr 23, 2013 by sreemathi.v

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