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Question Number 11822 by Joel576 last updated on 01/Apr/17

∫ ((tan x)/(1 + sin x)) dx

$$\int\:\frac{\mathrm{tan}\:{x}}{\mathrm{1}\:+\:\mathrm{sin}\:{x}}\:{dx} \\ $$

Answered by ajfour last updated on 01/Apr/17

let x=(π/2)−θ  I=∫((tan x)/(1+sin x))dx =∫((−dθ)/(tan θ (1+cos θ )))  =−∫(( 1−tan^2 ((θ/2)))/(2tan((θ/2))2cos^2 ((θ/2)))) dθ  let tan (θ/2)=t ⇒ dt= (1/2)sec^2 (θ/2)dθ  I=−∫((1−t^2 )/(2t))dt =(1/2)∫(t−(1/t))dt  I= (t^2 /4)−(1/2)ln ∣t∣+C  and as  t=tan (θ/2)=tan ((π/4)−(x/2))  I= (1/4)tan^2 ((π/4)−(x/2))−(1/2)ln ∣(π/4)−(x/2)∣+C .

$${let}\:{x}=\frac{\pi}{\mathrm{2}}−\theta \\ $$$${I}=\int\frac{\mathrm{tan}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}{dx}\:=\int\frac{−{d}\theta}{\mathrm{tan}\:\theta\:\left(\mathrm{1}+\mathrm{cos}\:\theta\:\right)} \\ $$$$=−\int\frac{\:\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2tan}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}\:{d}\theta \\ $$$${let}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}={t}\:\Rightarrow\:{dt}=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} \left(\theta/\mathrm{2}\right){d}\theta \\ $$$${I}=−\int\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left({t}−\frac{\mathrm{1}}{{t}}\right){dt} \\ $$$${I}=\:\frac{{t}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid{t}\mid+{C} \\ $$$${and}\:{as}\:\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right) \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\mid+{C}\:. \\ $$

Commented by Joel576 last updated on 02/Apr/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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