Mathematics

# Integrate  with respect to x : $\dfrac{\cos 2x - \cos 2 \alpha}{\cos x - \cos \alpha}$

##### SOLUTION
Given, $\dfrac{\cos 2x-\cos 2\alpha}{\cos x-\cos \alpha}$

$=\displaystyle\int \dfrac{(\cos 2x-\cos 2\alpha)}{(\cos x-\cos \alpha)}dx$

$=\displaystyle\int \dfrac{(2\cos ^2x-1)-(2\cos ^2\alpha -1)}{(\cos x-\cos\alpha)\cdot} dx$

$=\displaystyle\int \dfrac{(2\cos^ 2x-2\cos ^2\alpha)}{(\cos x-\cos \alpha)}dx$

$=\displaystyle\int \dfrac{2(\cos x-\cos\alpha)(\cos\alpha +\cos x)}{(\cos x-\cos \alpha)}dx$

$=\displaystyle\int 2(\cos\alpha +\cos x)dx$

$=2\sin x+2x\cdot \cos\alpha +c$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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