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Question Number 92960 by i jagooll last updated on 10/May/20

∫ ((sin^2 x dx)/((1+cos x)^2 ))

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

Answered by i jagooll last updated on 10/May/20

((1−cos^2 x)/((1+cos x)^2 )) = ((1−cos x)/(1+cos x)) = tan^2 ((x/2)) = sec^2 ((x/2))−1 ∫ ((sin^2 x dx)/((1+cos x)^2 )) = ∫ [sec^2 ((x/2))−1] dx = 2 tan ((x/2)) − x + c

$$\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\:=\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$$=\:\mathrm{sec}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{1} \\ $$$$\int\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}}{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }\:=\:\int\:\left[\mathrm{sec}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{1}\right]\:\mathrm{dx} \\ $$$$=\:\mathrm{2}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:−\:\mathrm{x}\:+\:\mathrm{c}\: \\ $$

Commented by john santu last updated on 10/May/20

��cheerr

Answered by $@ty@m123 last updated on 10/May/20

Alternative way: ((sin^2 x)/((1+cos x)^2 )) =((1−cos x)/(1+cos x))×((1−cos x)/(1−cos x)) =(((1−cos x)/(sin x)))^2 =(cosec x−cot x)^2 =cosec^2 x+cot^2 x−2cosec xcot x =2cosec^2 x−1−2cosec xcot x ∴∫(2cosec^2 x−1−2cosec xcot x)dx =−2cot x−x+cosec x+C

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

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