Question-10566

Question Number 10566 by krist last updated on 18/Feb/17

Answered by ridwan balatif last updated on 18/Feb/17

∫((cosx)/(1−cos^2 x))dx =∫((cosx)/(sin^2 x))dx =∫((cosx)/(sinx))×(1/(sinx))dx =∫cotanx×cosecxdx =−cosecx+C

$$\int\frac{\mathrm{cos}{x}}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int\frac{\mathrm{cos}{x}}{\mathrm{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int\frac{\mathrm{cos}{x}}{\mathrm{sin}{x}}×\frac{\mathrm{1}}{\mathrm{sin}{x}}{dx} \\ $$$$=\int\mathrm{cotan}{x}×\mathrm{cosec}{xdx} \\ $$$$=−\mathrm{cosec}{x}+\mathrm{C} \\ $$

Answered by mrW1 last updated on 18/Feb/17

∫((cos x)/(1−cos^2 x))dx =∫((cos x)/(sin^2 x))dx =∫(1/(sin^2 x))d(sin x) =−(1/(sin x))+C =−cosec x+C

$$\int\frac{\mathrm{cos}\:{x}}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{x}}{dx} \\ $$$$=\int\frac{\mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:{x}}{d}\left(\mathrm{sin}\:{x}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+{C} \\ $$$$=−\mathrm{cosec}\:{x}+{C} \\ $$

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Question-10566

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