Question
Evaluate: $\int\frac{\text{cos 2x}\text{ - cos }\alpha}{\text{cos x - cos }\alpha}\text{dx}.$
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Answer
Evaluate: $\text{I}=\int\frac{\text{cos 2x}\text{ - cos }\alpha}{\text{cos x - cos }\alpha}\text{dx}$
$=\int\frac{\text{(2cos}^{2}\text{x - 1)}-\text{(2 cos}^{2}\alpha- 1)}{\text{cos x - cos }\alpha}\text{dx}$
$=\int\frac{\text{2(cos}^{2}\text{x - cos}^{2}\alpha)}{\text{cos x - cos }\alpha}\text{dx}$ $=\int\frac{\text{2(cos x + cos }\alpha).\text{(cos x - cos }\alpha)}{\text{cos x - cos }\alpha}\text{dx}$
$=\text{2}\int\text{(cos x + cos }\alpha)\text{dx = 2}\int\text{cos x dx +2}\int\text{cos}\alpha\text{ dx}$
= 2 sin x + 2x cos $\alpha$ + C.
$=\int\frac{\text{(2cos}^{2}\text{x - 1)}-\text{(2 cos}^{2}\alpha- 1)}{\text{cos x - cos }\alpha}\text{dx}$
$=\int\frac{\text{2(cos}^{2}\text{x - cos}^{2}\alpha)}{\text{cos x - cos }\alpha}\text{dx}$ $=\int\frac{\text{2(cos x + cos }\alpha).\text{(cos x - cos }\alpha)}{\text{cos x - cos }\alpha}\text{dx}$
$=\text{2}\int\text{(cos x + cos }\alpha)\text{dx = 2}\int\text{cos x dx +2}\int\text{cos}\alpha\text{ dx}$
= 2 sin x + 2x cos $\alpha$ + C.
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