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Question Number 159870 by tounghoungko last updated on 21/Nov/21

∫ ((1−cot^2 x)/(1+sin x)) dx =?

$$\:\:\int\:\frac{\mathrm{1}−\mathrm{cot}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}\:{x}}\:{dx}\:=? \\ $$

Answered by Ar Brandon last updated on 21/Nov/21

I=∫((1−cot^2 x)/(1+sinx))dx=∫(((1−cot^2 x)(1−sinx))/(1−sin^2 x))dx =∫((1−sinx−cot^2 x+cot^2 xsinx)/(cos^2 x))dx =∫((1/(cos^2 x))−((sinx)/(cos^2 x))−(1/(sin^2 x))+(1/(sinx)))dx =tanx−(1/(cosx))+cotx+ln∣cosecx−cotx∣+C =((sinx−1)/(cosx))+((cosx)/(sinx))+ln∣((1−cosx)/(sinx))∣+C

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

Commented by tounghoungko last updated on 22/Nov/21

yes...very simple

$${yes}...{very}\:{simple} \\ $$

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