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Question Number 40675 by Raj Singh last updated on 26/Jul/18

Commented by math khazana by abdo last updated on 30/Jul/18

let I = ∫ (dx/(2sinx +cosx +3)) cha7gement tan((x/2))=t give I = ∫ (1/(2((2t)/(1+t^2 )) +((1−t^2 )/(1+t^2 )) +3)) ((2dt)/(1+t^2 )) = ∫ ((2dt)/(4t +1−t^2 +3+3t^2 )) = ∫ ((2dt)/(2t^2 +4t +4)) =∫ (dt/(t^2 +2t +2)) =∫ (dt/((t+1)^2 +1)) =_(t+1 =tanθ) ∫ (((1+tan^2 θ)dθ)/(1+tan^2 θ)) =θ +c =arctan(1+t) +c =arctan(1+tan((x/2))) +c .

$${let}\:{I}\:=\:\int\:\:\:\:\:\frac{{dx}}{\mathrm{2}{sinx}\:+{cosx}\:+\mathrm{3}}\:{cha}\mathrm{7}{gement} \\ $$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:\:\:=\:\int\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\mathrm{3}}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{4}{t}\:+\mathrm{1}−{t}^{\mathrm{2}} +\mathrm{3}+\mathrm{3}{t}^{\mathrm{2}} }\:=\:\int\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{4}{t}\:+\mathrm{4}} \\ $$$$=\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{2}}\:=\int\:\:\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=_{{t}+\mathrm{1}\:={tan}\theta} \:\:\int\:\:\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:=\theta\:+{c} \\ $$$$={arctan}\left(\mathrm{1}+{t}\right)\:+{c} \\ $$$$={arctan}\left(\mathrm{1}+{tan}\left(\frac{{x}}{\mathrm{2}}\right)\right)\:+{c}\:. \\ $$

Commented by math khazana by abdo last updated on 30/Jul/18

sir Raj i think the Q is prove not solve...

$${sir}\:{Raj}\:{i}\:{think}\:{the}\:{Q}\:{is}\:{prove}\:{not}\:{solve}... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

∫(dx/(2sinx+cosx+3)) let t=tan(x/2) dt=sec^2 (x/2)×(1/2)dx ((2dt)/(1+t^2 ))=dx ∫((2dt)/((1+t^2 )(((4t)/(1+t^2 ))+((1−t^2 )/(1+t^2 ))+3))) ∫((2dt)/(4t+1−t^2 +3+3t^2 )) ∫((2dt)/(2t^2 +4t+4)) ∫(dt/((t+1)^2 +1)) tan^(−1) (((t+1)/1))+c tan^(−1) (((1+tan(x/2))/1))+c

$$\int\frac{{dx}}{\mathrm{2}{sinx}+{cosx}+\mathrm{3}} \\ $$$${let}\:{t}={tan}\frac{{x}}{\mathrm{2}}\:\:\:{dt}={sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}{dx} \\ $$$$\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }={dx} \\ $$$$\int\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\frac{\mathrm{4}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{3}\right)} \\ $$$$\int\frac{\mathrm{2}{dt}}{\mathrm{4}{t}+\mathrm{1}−{t}^{\mathrm{2}} +\mathrm{3}+\mathrm{3}{t}^{\mathrm{2}} } \\ $$$$\int\frac{\mathrm{2}{dt}}{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{4}} \\ $$$$\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{{t}+\mathrm{1}}{\mathrm{1}}\right)+{c} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}}\right)+{c} \\ $$$$ \\ $$$$ \\ $$

Commented by Raj Singh last updated on 26/Jul/18

ttthhannks

$${ttthhannks}\: \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

its ok...

$${its}\:{ok}... \\ $$

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