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Question Number 46720 by Necxx last updated on 30/Oct/18
Commented by Necxx last updated on 30/Oct/18
please help with no. 15 cos i got it as −sin^(−1) ((1/x))− ((√(x^2 +1))/x) +c
$${please}\:{help}\:{with}\:{no}.\:\mathrm{15}\:{cos}\:{i}\:{got}\:{it} \\ $$$${as} \\ $$$$−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)−\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}}\:+{c} \\ $$
Commented by maxmathsup by imad last updated on 30/Oct/18
let A =∫ (1/x^2 )(√((x−1)/(x+1)))dx changement ((x−1)/(x+1)) =t^2 give x−1=t^2 x +t^2 ⇒ (1−t^2 )x=1+t^2 ⇒x=((1+t^2 )/(1−t^2 )) ⇒ dx =((2t(1−t^2 )+2t(1+t^2 ))/((1−t^2 )^2 ))dt =((2t −2t^3 +2t +2t^3 )/((1−t^2 )^2 ))dt =((4t)/((1−t^2 )^2 ))dt ⇒A=∫ (((1−t^2 )^2 )/((1+t^2 )^2 )) t ((4t)/((1−t^2 )^2 ))dt =4 ∫ (t^2 /((1+t^2 )^2 ))dt =4 ∫ ((1+t^2 −1)/((1+t^2 )^2 )) dt =4 ∫ (dt/(1+t^2 )) −4 ∫ (dt/((1+t^2 )^2 )) =4 arctan(t)−4 ∫ (dt/((1+t^2 )^2 )) changement t=tanθ give ∫ (dt/((1+t^2 )^2 )) = ∫ ((1+tan^2 θ)/((1+tan^2 θ)^2 ))dθ =∫ (dθ/(1+tan^2 θ)) =∫ cos^2 θ dθ = =(1/2) ∫ (1+cos(2θ))dθ =(θ/2) +(1/4)sin(2θ) =((arctan(t))/2) +(1/4)sin(2arctant) but sin(2arctant)=2 sin(arctant)cos(arctant) =2 (t/( (√(1+t^2 )))) .(1/( (√(1+t^2 )))) =((2t)/(1+t^2 )) ⇒ ∫ (dt/((1+t^2 )^2 )) =((arctan(t))/2) +(t/(2(1+t^2 ))) ⇒ A =2arctan(t)−((2t)/(1+t^2 )) +c =2 arctan((√((x−1)/(x+1))))−((2(√((x−1)/(x+1))))/(1+((x−1)/(x+1)))) +c =2 arctan((√((x−1)/(x+1))))−2 (√((x−1)/(x+1)))((x+1)/(2x)) +c A=2 arctan((√((x−1)/(x+1)))) +((√(x^2 −1))/x) +c
$${let}\:{A}\:=\int\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}{dx}\:{changement}\:\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\:={t}^{\mathrm{2}} \:{give}\:{x}−\mathrm{1}={t}^{\mathrm{2}} {x}\:+{t}^{\mathrm{2}} \:\Rightarrow \\ $$$$\left(\mathrm{1}−{t}^{\mathrm{2}} \right){x}=\mathrm{1}+{t}^{\mathrm{2}} \:\Rightarrow{x}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\:\Rightarrow\:{dx}\:=\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+\mathrm{2}{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{2}{t}\:−\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{2}{t}\:+\mathrm{2}{t}^{\mathrm{3}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=\frac{\mathrm{4}{t}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:\Rightarrow{A}=\int\:\:\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{t}\:\:\frac{\mathrm{4}{t}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{4}\:\int\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=\mathrm{4}\:\int\:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:=\mathrm{4}\:\int\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:−\mathrm{4}\:\int\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:{arctan}\left({t}\right)−\mathrm{4}\:\int\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{changement}\:{t}={tan}\theta\:{give} \\ $$$$\int\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:\int\:\:\frac{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }{d}\theta\:=\int\:\:\frac{{d}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:=\int\:{cos}^{\mathrm{2}} \theta\:{d}\theta\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)\right){d}\theta\:=\frac{\theta}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}\theta\right)\:=\frac{{arctan}\left({t}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arctant}\right)\:{but} \\ $$$${sin}\left(\mathrm{2}{arctant}\right)=\mathrm{2}\:{sin}\left({arctant}\right){cos}\left({arctant}\right) \\ $$$$=\mathrm{2}\:\frac{{t}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow\:\int\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{{arctan}\left({t}\right)}{\mathrm{2}}\:+\frac{{t}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${A}\:=\mathrm{2}{arctan}\left({t}\right)−\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+{c} \\ $$$$=\mathrm{2}\:{arctan}\left(\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\right)−\frac{\mathrm{2}\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}}{\mathrm{1}+\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:+{c} \\ $$$$=\mathrm{2}\:{arctan}\left(\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\right)−\mathrm{2}\:\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\frac{{x}+\mathrm{1}}{\mathrm{2}{x}}\:+{c} \\ $$$${A}=\mathrm{2}\:{arctan}\left(\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\right)\:+\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}}\:+{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Necxx last updated on 30/Oct/18
thank you so much sir . The issue now is that this answer is not in the option. Does it mean the options are all wrong?
$${thank}\:{you}\:{so}\:{much}\:{sir}\:.\: \\ $$$${The}\:{issue}\:{now}\:{is}\:{that}\:{this}\:{answer} \\ $$$${is}\:{not}\:{in}\:{the}\:{option}.\:{Does}\:{it}\:{mean} \\ $$$${the}\:{options}\:{are}\:{all}\:{wrong}? \\ $$
Commented by maxmathsup by imad last updated on 30/Oct/18
yes there is some error in the given answers but always you must trust yourself and your method...
$${yes}\:{there}\:{is}\:{some}\:{error}\:{in}\:{the}\:{given}\:{answers}\:{but}\:{always}\:{you}\:{must}\:{trust} \\ $$$${yourself}\:{and}\:{your}\:{method}… \\ $$
Commented by maxmathsup by imad last updated on 30/Oct/18
16) let I = ∫ (dx/( (√(x−x^2 )))) ⇒I =∫ (dx/( (√(−(x^2 −x))))) =∫ (dx/( (√(−(x^2 −2(1/2)x+(1/4)−(1/4)))))) = ∫ (dx/( (√((1/4)−(x−(1/2))^2 )))) =_(x−(1/2)=(1/2)sint) ∫ (1/(2.(1/2)cost)) cost dt = ∫ dt =t +c = arcsin(2x−1) +c so option (B) is the answer.
$$\left.\mathrm{16}\right)\:{let}\:\:{I}\:=\:\int\:\frac{{dx}}{\:\sqrt{{x}−{x}^{\mathrm{2}} }}\:\Rightarrow{I}\:=\int\:\:\frac{{dx}}{\:\sqrt{−\left({x}^{\mathrm{2}} −{x}\right)}}\:=\int\:\:\frac{{dx}}{\:\sqrt{−\left({x}^{\mathrm{2}} −\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\right)}} \\ $$$$=\:\int\:\:\:\frac{{dx}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }}\:=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{sint}} \:\:\:\int\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}{cost}}\:{cost}\:{dt} \\ $$$$=\:\int\:{dt}\:={t}\:+{c}\:=\:{arcsin}\left(\mathrm{2}{x}−\mathrm{1}\right)\:+{c}\:\:{so}\:{option}\:\left({B}\right)\:{is}\:{the}\:{answer}. \\ $$
Commented by Necxx last updated on 30/Oct/18
wow.... i solved this and got A as my answer
$${wow}….\:{i}\:{solved}\:{this}\:{and}\:{got}\:{A}\:{as} \\ $$$${my}\:{answer} \\ $$$$ \\ $$

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