1-1-sinx-dx-

Question Number 29885 by sinx last updated on 13/Feb/18

$$\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}{x}}{dx}=? \\ $$

Commented by MJS last updated on 14/Feb/18

=tanx−(1/(cosx)) ...sorry, can′t show the way...

$$=\mathrm{tan}{x}−\frac{\mathrm{1}}{\mathrm{cos}{x}} \\ $$$$…\mathrm{sorry},\:\mathrm{can}'\mathrm{t}\:\mathrm{show}\:\mathrm{the}\:\mathrm{way}… \\ $$

Commented by abdo imad last updated on 14/Feb/18

f(x)=((tanx cosx−1)/(cosx))= ((sinx−1)/(cosx)) ⇒ f^′ (x)= ((cos^2 x +(sinx−1)sinx)/(cos^2 x)) = ((1−sinx)/(1−sin^2 x)) =(1/(1+sinx)) so ∫ (dx/(1+sinx))= tanx −(1/(cosx)) +k .

$${f}\left({x}\right)=\frac{{tanx}\:{cosx}−\mathrm{1}}{{cosx}}=\:\frac{{sinx}−\mathrm{1}}{{cosx}}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\:\frac{{cos}^{\mathrm{2}} {x}\:+\left({sinx}−\mathrm{1}\right){sinx}}{{cos}^{\mathrm{2}} {x}}\:=\:\frac{\mathrm{1}−{sinx}}{\mathrm{1}−{sin}^{\mathrm{2}} {x}}\:=\frac{\mathrm{1}}{\mathrm{1}+{sinx}}\:{so} \\ $$$$\int\:\:\frac{{dx}}{\mathrm{1}+{sinx}}=\:{tanx}\:−\frac{\mathrm{1}}{{cosx}}\:+{k}\:. \\ $$

Answered by Joel578 last updated on 13/Feb/18

I = ∫ (1/(1 + sin x)) . ((1 − sin x)/(1 − sin x)) dx = ∫ ((1 − sin x)/(1 − sin^2 x)) dx = ∫ ((1 − sin x)/(cos^2 x)) dx = ∫ ((1/(cos^2 x)) − tan x sec x) dx = ∫ sec^2 x dx − ∫ tan x sec x dx

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

Commented by abdo imad last updated on 13/Feb/18

let put I=∫ (dx/(1+sinx)) the ch.tan((x/2))=t give I = ∫ (1/(1+((2t)/(1+t^2 )))) ((2dt)/(1+t^2 ))= ∫ ((2dt)/(1+t^2 +2t)) = ∫ (dt/((t+1)^2 )) =−(1/(t+1)) +k =((−1)/(1+tan((x/2)))) +k

$${let}\:{put}\:{I}=\int\:\:\:\frac{{dx}}{\mathrm{1}+{sinx}}\:{the}\:{ch}.{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\:\int\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\:\:\int\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{2}{t}}\:=\:\int\:\:\:\:\:\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{{t}+\mathrm{1}}\:+{k}\:\:=\frac{−\mathrm{1}}{\mathrm{1}+{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\:+{k} \\ $$

Commented by NECx last updated on 13/Feb/18

$${hmmm} \\ $$$$ \\ $$

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