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cos-x-cos-2x-1-cos-x-dx-




Question Number 27815 by goswamisubhabrata007@gmail.com last updated on 15/Jan/18
∫((cos x−cos 2x)/(1−cos x))dx
$$\int\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{cos}\:\mathrm{2x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}\mathrm{dx} \\ $$
Answered by ajfour last updated on 15/Jan/18
∫((cos x−2cos^2 x+1)/(1−cos x)) dx  =∫(((2cos x−2cos^2 x+1−cos x))/((1−cos x))) dx  =∫(((1−cos x)(1+2cos x))/((1−cos x))) dx  =∫(1+2cos x)dx  =x+2sin x+c .
$$\int\frac{\mathrm{cos}\:{x}−\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{1}}{\mathrm{1}−\mathrm{cos}\:{x}}\:{dx} \\ $$$$=\int\frac{\left(\mathrm{2cos}\:{x}−\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{1}−\mathrm{cos}\:{x}\right)}{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}\:{dx} \\ $$$$=\int\frac{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)\left(\mathrm{1}+\mathrm{2cos}\:{x}\right)}{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}\:{dx} \\ $$$$=\int\left(\mathrm{1}+\mathrm{2cos}\:{x}\right){dx} \\ $$$$={x}+\mathrm{2sin}\:{x}+{c}\:. \\ $$

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