First slide

Introduction to integration

Question

$\int \frac{\cos 2 x - \cos 2 α}{\cos x - \cos α} dx$ is equal to

Moderate

A

$2 \sin x + 2 xcos α + C$
B

$\sin x + xcos α + C$
C

$2 \sin x + xcos x + C$
D

None of these

Solution

$\begin{array}{l} \int \frac{\cos 2 x - \cos 2 α}{\cos x - \cos α} dx \\ = \int \frac{(2 \cos^{2} x - 1) - (2 \cos^{2} α - 1)}{(\cos x - \cos α)} dx [∵ \cos 2 x = 2 \cos^{2} x - 1] \\ = \int \frac{2 \cos^{2} x - 1 - 2 \cos^{2} α + 1}{(\cos x - \cos α)} dx \\ = \int \frac{2 (\cos^{2} x - \cos^{2} α)}{(\cos x - \cos α)} dx \\ = 2 \int \frac{(\cos x - \cos α) (\cos x + \cos α)}{(\cos x - \cos α)} dx \\ [∵ a^{2} - b^{2} = (a + b) (a - b)] \\ = 2 [\int \cos xdx + \cos α \int 1 dx] \\ = 2 [\sin x + \cos α \cdot x] + C \\ = 2 \sin x + 2 xcos α + C \end{array}$

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