Question-81889

Question Number 81889 by rajesh4661kumar@gmail.com last updated on 16/Feb/20

Answered by john santu last updated on 16/Feb/20

let (√x) = sin t ⇒x = sin^2 t dx = 2sin t cos t dt ⇒I = ∫ (√(((1−sin t)/(1+sin t)) )) (2sin t cos t ) dt = ∫ (((1−sin t)2sin tcos t)/(cos t)) dt = ∫ (1−sin t)2sin t dt = ∫ 2sin t−2((1/2)−(1/2)cos 2t) dt = −2cos t−t+(1/2)sin 2t +c = −2(√(1−x)) −sin^(−1) ((√x))+(√(x−x^2 )) +c

$${let}\:\sqrt{{x}}\:=\:\mathrm{sin}\:{t}\:\Rightarrow{x}\:=\:\mathrm{sin}\:^{\mathrm{2}} {t} \\ $$$${dx}\:=\:\mathrm{2sin}\:{t}\:\mathrm{cos}\:{t}\:{dt}\: \\ $$$$\Rightarrow{I}\:=\:\int\:\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:{t}}{\mathrm{1}+\mathrm{sin}\:{t}}\:}\:\left(\mathrm{2sin}\:{t}\:\mathrm{cos}\:{t}\:\right)\:{dt} \\ $$$$=\:\int\:\frac{\left(\mathrm{1}−\mathrm{sin}\:{t}\right)\mathrm{2sin}\:{t}\mathrm{cos}\:{t}}{\mathrm{cos}\:{t}}\:{dt} \\ $$$$=\:\int\:\left(\mathrm{1}−\mathrm{sin}\:{t}\right)\mathrm{2sin}\:{t}\:{dt} \\ $$$$=\:\int\:\mathrm{2sin}\:{t}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{t}\right)\:{dt} \\ $$$$=\:−\mathrm{2cos}\:{t}−{t}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{t}\:+{c} \\ $$$$=\:−\mathrm{2}\sqrt{\mathrm{1}−{x}}\:−\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{{x}}\right)+\sqrt{{x}−{x}^{\mathrm{2}} }\:+{c} \\ $$

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Question-81889

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