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Question Number 43913 by maxmathsup by imad last updated on 17/Sep/18
find ∫ (dx/( (√(1−cosx)) +(√(1+cosx))))
$${find}\:\int\:\:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{cosx}}\:+\sqrt{\mathrm{1}+{cosx}}} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 18/Sep/18
I =∫ (dx/( (√(2sin^2 ((x/2))))+(√(2cos^2 ((x/2)))))) =(1/( (√2))) ∫ (dx/(sin((x/2))+cos((x/2)))) =_((x/2)=t) (1/( (√2)))∫ ((2dt)/(sint +cost)) dt =(1/( (√2)))∫ ((2dt)/( (√2)cos(t−(π/4))))dt=_(t−(π/4)=u) ∫ (du/(cosu)) =_(tan((u/2))=α) ∫ (1/((1−α^2 )/(1+α^2 ))) ((2dα)/(1+α^2 )) =2∫ (dα/(1−α^2 )) = ∫ {(1/(1−α)) +(1/(1+α))}dα =ln∣((1+α)/(1−α))∣ =ln∣((1+tan((u/2)))/(1−tan((u/2))))∣ =ln∣tan((u/2)+(π/4))∣ +c =ln∣tan((t/2) −(π/8) +(π/4))∣ +c = ln∣tan((t/2)+(π/8))∣ +c
$${I}\:=\int\:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}+\sqrt{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\:\:\frac{{dx}}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)+{cos}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$=_{\frac{{x}}{\mathrm{2}}={t}} \:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\:\:\:\:\frac{\mathrm{2}{dt}}{{sint}\:+{cost}}\:{dt}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\:\:\:\frac{\mathrm{2}{dt}}{\:\sqrt{\mathrm{2}}{cos}\left({t}−\frac{\pi}{\mathrm{4}}\right)}{dt}=_{{t}−\frac{\pi}{\mathrm{4}}={u}} \:\int\:\:\frac{{du}}{{cosu}} \\ $$$$=_{{tan}\left(\frac{{u}}{\mathrm{2}}\right)=\alpha} \:\:\:\:\:\:\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−\alpha^{\mathrm{2}} }{\mathrm{1}+\alpha^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }\:=\mathrm{2}\int\:\:\:\frac{{d}\alpha}{\mathrm{1}−\alpha^{\mathrm{2}} }\:=\:\int\:\:\left\{\frac{\mathrm{1}}{\mathrm{1}−\alpha}\:+\frac{\mathrm{1}}{\mathrm{1}+\alpha}\right\}{d}\alpha \\ $$$$={ln}\mid\frac{\mathrm{1}+\alpha}{\mathrm{1}−\alpha}\mid\:={ln}\mid\frac{\mathrm{1}+{tan}\left(\frac{{u}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{u}}{\mathrm{2}}\right)}\mid\:={ln}\mid{tan}\left(\frac{{u}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mid\:+{c} \\ $$$$={ln}\mid{tan}\left(\frac{{t}}{\mathrm{2}}\:−\frac{\pi}{\mathrm{8}}\:+\frac{\pi}{\mathrm{4}}\right)\mid\:+{c}\:=\:{ln}\mid{tan}\left(\frac{{t}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)\mid\:+{c}\: \\ $$
Commented by maxmathsup by imad last updated on 18/Sep/18
but t =(x/2) ⇒ I =ln∣tan((x/4)+(π/8))∣ +c .
$${but}\:{t}\:=\frac{{x}}{\mathrm{2}}\:\Rightarrow\:{I}\:={ln}\mid{tan}\left(\frac{{x}}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}\right)\mid\:+{c}\:. \\ $$
Commented by math1967 last updated on 18/Sep/18
It is simplest form of cosec((x/2)+(π/4))−cot((x/2)+(π/4)) sir is it not?
$${It}\:{is}\:{simplest}\:{form}\:{of}\: \\ $$$${cosec}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)−{cot}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\:{sir} \\ $$$${is}\:{it}\:{not}? \\ $$
Commented by maxmathsup by imad last updated on 18/Sep/18
perhaps because i dont use the symbol cosec and sec...
$${perhaps}\:{because}\:{i}\:{dont}\:{use}\:{the}\:{symbol}\:\:{cosec}\:{and}\:{sec}… \\ $$
Answered by math1967 last updated on 18/Sep/18
∫(dx/( (√(2sin^2 (x/2))) +(√(2cos^2 (x/2))))) ∫(dx/( (√2)sin(x/2)+(√2)cos(x/2))) ∫(dx/(2((1/( (√2)))sin(x/2)+(1/( (√2)))cos(x/2)))) ∫(dx/(2(sin(x/2)cos(π/4)+cos(x/2)sin(π/4)))) (1/2)∫ (dx/(sin((x/2)+(π/4)))) (1/2)∫cosec((x/2)+(π/4))dx ∫cosec((x/2)+(π/4))d((x/2)+(π/4)) ln[cosec((x/2)+(π/4))−cot((x/2)+(π/4))] +c
$$\int\frac{{dx}}{\:\sqrt{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}\:+\sqrt{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}} \\ $$$$\int\frac{{dx}}{\:\sqrt{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}+\sqrt{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}} \\ $$$$\int\frac{{dx}}{\mathrm{2}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sin}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{cos}\frac{{x}}{\mathrm{2}}\right)} \\ $$$$\int\frac{{dx}}{\mathrm{2}\left({sin}\frac{{x}}{\mathrm{2}}{cos}\frac{\pi}{\mathrm{4}}+{cos}\frac{{x}}{\mathrm{2}}{sin}\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{dx}}{{sin}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int{cosec}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right){dx} \\ $$$$\int{cosec}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right){d}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right) \\ $$$${ln}\left[{cosec}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)−{cot}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right]\:+{c} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Sep/18
∫(dx/( (√2) sin(x/2)+(√2) cos(x/2))) ∫(dx/(2×((1/( (√2) ))sin(x/2)+(1/( (√2) ))cos(x/2)))) (1/2)∫cosec((x/2)+(Π/4))dx [∫cosecx=lntan(x/2)] (1/2)ln{tan((x/4)+(Π/8))}×(1/(1/2)) ln{tan((x/4)+(Π/8))}+c
$$\int\frac{{dx}}{\:\sqrt{\mathrm{2}}\:{sin}\frac{{x}}{\mathrm{2}}+\sqrt{\mathrm{2}}\:{cos}\frac{{x}}{\mathrm{2}}} \\ $$$$\int\frac{{dx}}{\mathrm{2}×\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}{sin}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}{cos}\frac{{x}}{\mathrm{2}}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int{cosec}\left(\frac{{x}}{\mathrm{2}}+\frac{\Pi}{\mathrm{4}}\right){dx}\:\:\left[\int{cosecx}={lntan}\frac{{x}}{\mathrm{2}}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{{tan}\left(\frac{{x}}{\mathrm{4}}+\frac{\Pi}{\mathrm{8}}\right)\right\}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${ln}\left\{{tan}\left(\frac{{x}}{\mathrm{4}}+\frac{\Pi}{\mathrm{8}}\right)\right\}+{c} \\ $$

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