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\( \int \frac{\cos x-\cos 2 x}{1-\cos x} d x \) |
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Answer» ∵ \(\frac{cos\,x-cos\,2x}{1-cos\,x}=\frac{cos\,x-(2cos^2x-1)}{1-cos\,x}\) (∵ cos 2x = 2 cos2x - 1) \(=\frac{-2cos^2x+cos\,x+1)}{1-cos\,x}\) \(=\frac{-2cos^2x+2\,cos\,x-cos\,x+1)}{1-cos\,x}\) \(=\frac{2\,cos\,x(1-cos\,x)+(1-cos\,x)}{1-cos\,x}\) \(=\frac{(1-cos\,x)(2\,cos\,x+1)}{1-cos\,x}\) = 2 cos x+1 ∴ ∫\(\frac{cos\,x-cos\,2x}{1-cos\,x}dx\) = ∫(2 cos x+1)dx = 2∫cos x dx + ∫dx = 2 sin x+x+c |
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