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Question Number 112313 by mathdave last updated on 07/Sep/20

solve ∫_0 ^1 ((lnx)/(√(1+x^2 )))dx

$${solve} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$

Commented by mohammad17 last updated on 07/Sep/20

set:x=tan(r)⇒dx=sec^2 (r)dr x=0⇒r=0 , x=1⇒r=(π/4) ∫_0 ^( (π/4)) sec^3 (r)dr by using the rule ∫sec^n (x)dx=((sec^(n−1) (x))/((n−1)csc(x)))+(((n−2))/((n−1)))∫sec^(n−2) (x)dx ∴∫_0 ^( (π/4)) sec^3 (r)dr=(((sec^2 (r)sin(r))/2))_0 ^( (π/4)) +(1/2)∫_0 ^( (π/4)) sec(r)dr Note:∫sec(r)dr=ln∣sec(r)+tan(r)∣+C ∴∫_0 ^( 1) (√(1+x^2 ))dx=(1/2)((√2) )+(1/2)ln∣(√2)+1∣=(((√2)+ln∣1+(√2)∣)/2)

$${set}:{x}={tan}\left({r}\right)\Rightarrow{dx}={sec}^{\mathrm{2}} \left({r}\right){dr} \\ $$$${x}=\mathrm{0}\Rightarrow{r}=\mathrm{0}\:\:,\:\:{x}=\mathrm{1}\Rightarrow{r}=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {sec}^{\mathrm{3}} \left({r}\right){dr}\: \\ $$$$ \\ $$$${by}\:{using}\:{the}\:{rule} \\ $$$$ \\ $$$$\int{sec}^{{n}} \left({x}\right){dx}=\frac{{sec}^{{n}−\mathrm{1}} \left({x}\right)}{\left({n}−\mathrm{1}\right){csc}\left({x}\right)}+\frac{\left({n}−\mathrm{2}\right)}{\left({n}−\mathrm{1}\right)}\int{sec}^{{n}−\mathrm{2}} \left({x}\right){dx} \\ $$$$ \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {sec}^{\mathrm{3}} \left({r}\right){dr}=\left(\frac{{sec}^{\mathrm{2}} \left({r}\right){sin}\left({r}\right)}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\:\frac{\pi}{\mathrm{4}}} {\right)}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {sec}\left({r}\right){dr} \\ $$$$ \\ $$$${Note}:\int{sec}\left({r}\right){dr}={ln}\mid{sec}\left({r}\right)+{tan}\left({r}\right)\mid+{C} \\ $$$$ \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{2}}\:\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\sqrt{\mathrm{2}}+\mathrm{1}\mid=\frac{\sqrt{\mathrm{2}}+{ln}\mid\mathrm{1}+\sqrt{\mathrm{2}}\mid}{\mathrm{2}} \\ $$

Commented by mathdave last updated on 07/Sep/20

am so so sorry sir′s to change the phase of the question.the one i type before i eluded to include (lnx) ontop of (√(1+x^2 )).all works u guys did initially was all correct i cant deny that fact.u guys re great mathematicians

$${am}\:{so}\:{so}\:{sorry}\:{sir}'{s}\:{to}\:{change}\:{the}\:{phase}\:{of}\:{the} \\ $$$${question}.{the}\:{one}\:{i}\:{type}\:{before}\:{i}\:{eluded} \\ $$$${to}\:{include}\:\left(\mathrm{ln}{x}\right)\:{ontop}\:{of}\:\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }.{all}\:{works} \\ $$$${u}\:{guys}\:{did}\:{initially}\:{was}\:{all}\:{correct}\:{i} \\ $$$${cant}\:{deny}\:{that}\:{fact}.{u}\:{guys}\:{re}\:{great} \\ $$$${mathematicians} \\ $$

Answered by Dwaipayan Shikari last updated on 07/Sep/20

[(x/2)(√(1+x^2 ))+(1/2) log(x+(√(x^2 +1)))]_0 ^1 =(1/( (√2)))+(1/2)log(1+(√2))

$$\left[\frac{{x}}{\mathrm{2}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\:{log}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$

Answered by MJS_new last updated on 07/Sep/20

∫_0 ^1 (√(x^2 +1))dx= [t=x+(√(x^2 +1)) → dx=((√(x^2 +1))/(x+(√(x^2 +1))))dt] =∫_1 ^(1+(√2)) ((t/4)+(1/(2t))+(1/(4t^3 )))dt= =[(t^2 /8)+((ln t)/2)−(1/(8t^2 ))]_1 ^(1+(√2)) =(((√2)+ln (1+(√2)))/2)

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dt}\right] \\ $$$$=\underset{\mathrm{1}} {\overset{\mathrm{1}+\sqrt{\mathrm{2}}} {\int}}\left(\frac{{t}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}{t}}+\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{3}} }\right){dt}= \\ $$$$=\left[\frac{{t}^{\mathrm{2}} }{\mathrm{8}}+\frac{\mathrm{ln}\:{t}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{8}{t}^{\mathrm{2}} }\right]_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} =\frac{\sqrt{\mathrm{2}}+\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}} \\ $$

Answered by mathdave last updated on 07/Sep/20

solution to ∫_0 ^1 ((lnx)/(√(1+x^2 )))dx let x=sinhθ,dx=coshθdθ and sinh^(−1) (x)=ln(x+(√(1+x^2 ))) I=∫_(sinh^(−1) (0)) ^(sinh^(−1) (1)) ((ln(sinhθ))/(coshθ))dθ=∫_0 ^(sinh^(−1) (1)) ln(sinhθ)dθ but sinhθ=((e^θ −e^(−θ) )/2) I=∫_0 ^(ln(1+(√2))) ln(e^θ −e^(−θ) )dθ−∫_0 ^(ln(1+(√2))) ln(2)dθ I=∫_0 ^(ln(1+(√2))) ln(e^θ )dθ+∫_0 ^(ln(1+(√2))) ln(1−e^(−θ) )dθ−∫_0 ^(ln(1+(√2))) ln(2)dθ I=((ln^2 (1+(√2)))/2)−∫_0 ^(ln(1+(√2))) Σ_(k=0) ^∞ (e^(−2(k+1)θ) /(k+1))dθ−ln(2)ln(1+(√2)) I=((ln^2 (1+(√2)))/2)+(1/2)Σ_(k=0) ^∞ (1/(k+1))[((e^(−2(k+1)ln(1+(√2))) −1)/(k+1))]−ln(2)ln(1+(√2)) I=((ln^2 (1+(√2)))/2)+(1/2)Σ_(k=0) ^∞ (((1+(√2))^(−2(k+1)_ ) )/((k+1)^2 ))−(1/2)Σ_(k=0) ^∞ (1/((k+1)^2 ))−ln(2)ln(1+(√2)) I=((ln^2 (1+(√2)))/2)+(1/2)Σ_(k=1) ^∞ (1/(((1+(√2))^(2k) )/k^2 ))−(1/2)ζ2)−ln(2)ln(1+(√2)) but Li_2 (x)=Σ_(k=1) ^∞ (x^k /k^2 ) I=((ln^2 (1+(√2)))/2)+(1/2)Li_2 ((1/((1+(√2))^2 )))−(1/2)((π^2 /6))−ln(2)ln(1+(√2)) I=((ln^2 (1+(√2)))/2)+(1/2)Li_2 ((1/(3+2(√2))))−(π^2 /(12))−ln(2)ln(1+(√2)) I=((ln^2 (1+(√2)))/2)+(1/2)Li_2 (3−2(√2))−(π^2 /(12))−ln(2)ln(1+(√2)) I=((ln^2 (1+(√2)))/2)−(π^2 /(12))−ln(2)ln(1+(√2))+(1/2)Li_2 (3−2(√2)) ∵∫_0 ^1 ((lnx)/(√(1+x^2 )))dx=((ln^2 (1+(√2)))/2)−(π^2 /(12))−ln(2)ln(1+(√2))+(1/2)Li_2 (3−2(√2)) by mathdave(07/09/2020)

$${solution}\:{to}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$$${let}\:{x}=\mathrm{sinh}\theta,{dx}=\mathrm{cosh}\theta{d}\theta\:\:{and} \\ $$$$\mathrm{sinh}^{−\mathrm{1}} \left({x}\right)=\mathrm{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\ $$$${I}=\int_{\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{0}\right)} ^{\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{1}\right)} \frac{\mathrm{ln}\left(\mathrm{sinh}\theta\right)}{\mathrm{cosh}\theta}{d}\theta=\int_{\mathrm{0}} ^{\mathrm{sinh}^{−\mathrm{1}} \left(\mathrm{1}\right)} \mathrm{ln}\left(\mathrm{sinh}\theta\right){d}\theta \\ $$$${but}\:\mathrm{sinh}\theta=\frac{{e}^{\theta} −{e}^{−\theta} }{\mathrm{2}} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \mathrm{ln}\left({e}^{\theta} −{e}^{−\theta} \right){d}\theta−\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \mathrm{ln}\left(\mathrm{2}\right){d}\theta \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \mathrm{ln}\left({e}^{\theta} \right){d}\theta+\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \mathrm{ln}\left(\mathrm{1}−{e}^{−\theta} \right){d}\theta−\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \mathrm{ln}\left(\mathrm{2}\right){d}\theta \\ $$$${I}=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{−\mathrm{2}\left({k}+\mathrm{1}\right)\theta} }{{k}+\mathrm{1}}{d}\theta−\mathrm{ln}\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$${I}=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{1}}\left[\frac{{e}^{−\mathrm{2}\left({k}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} −\mathrm{1}}{{k}+\mathrm{1}}\right]−\mathrm{ln}\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$${I}=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{−\mathrm{2}\left({k}+\mathrm{1}\right)_{} } }{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{ln}\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$$\left.{I}=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}{k}} }{{k}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\mathrm{2}}\zeta\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$${but}\:{Li}_{\mathrm{2}} \left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{k}} }{{k}^{\mathrm{2}} } \\ $$$${I}=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right)−\mathrm{ln}\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$${I}=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\mathrm{ln}\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$${I}=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\mathrm{ln}\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$${I}=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\mathrm{ln}\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\because\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\mathrm{ln}\left(\mathrm{2}\right)\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$${by}\:{mathdave}\left(\mathrm{07}/\mathrm{09}/\mathrm{2020}\right) \\ $$

Commented by Tinku Tara last updated on 08/Sep/20

mathdave, please stop using abusive language on this forum immediately.

$$\mathrm{mathdave},\:\mathrm{please}\:\mathrm{stop}\:\mathrm{using}\:\mathrm{abusive} \\ $$$$\mathrm{language}\:\mathrm{on}\:\mathrm{this}\:\mathrm{forum}\:\mathrm{immediately}. \\ $$

Commented by Her_Majesty last updated on 07/Sep/20

∫(√(1+x^2 ))dx with x=sinht leads to ∫cosh^2 t dt =(1/2)(t+cosht sinht)= =(1/2)(sinh^(−1) x+x(√(x^2 +1)))+C no need for such a long tour through the lowlands of math

$$\int\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{with}\:{x}={sinht}\:{leads}\:{to} \\ $$$$\int{cosh}^{\mathrm{2}} {t}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left({t}+{cosht}\:{sinht}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({sinh}^{−\mathrm{1}} {x}+{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)+{C} \\ $$$${no}\:{need}\:{for}\:{such}\:{a}\:{long}\:{tour}\:{through}\:{the} \\ $$$${lowlands}\:{of}\:{math} \\ $$

Commented by MJS_new last updated on 07/Sep/20

you cannot hold it back it seems. sad.

$$\mathrm{you}\:\mathrm{cannot}\:\mathrm{hold}\:\mathrm{it}\:\mathrm{back}\:\mathrm{it}\:\mathrm{seems}.\:\mathrm{sad}. \\ $$

Commented by MJS_new last updated on 07/Sep/20

and you too cannot hold it back. the two of you are perfect mirrors for one another.

$$\mathrm{and}\:\mathrm{you}\:\mathrm{too}\:\mathrm{cannot}\:\mathrm{hold}\:\mathrm{it}\:\mathrm{back}. \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{of}\:\mathrm{you}\:\mathrm{are}\:\mathrm{perfect}\:\mathrm{mirrors}\:\mathrm{for}\:\mathrm{one} \\ $$$$\mathrm{another}. \\ $$

Commented by mathdave last updated on 07/Sep/20

nt that i cant hold back but she is doing has if she monopoly of knowledge

$${nt}\:{that}\:{i}\:{cant}\:{hold}\:{back}\:{but}\:{she}\:{is}\:{doing} \\ $$$${has}\:{if}\:{she}\:{monopoly}\:{of}\:{knowledge} \\ $$

Commented by MJS_new last updated on 07/Sep/20

Mr Mathdave, it′s exactly the same with you. scroll down and read your own comments!

$$\mathrm{Mr}\:\mathrm{Mathdave},\:\mathrm{it}'\mathrm{s}\:\mathrm{exactly}\:\mathrm{the}\:\mathrm{same}\:\mathrm{with} \\ $$$$\mathrm{you}.\:\mathrm{scroll}\:\mathrm{down}\:\mathrm{and}\:\mathrm{read}\:\mathrm{your}\:\mathrm{own}\:\mathrm{comments}! \\ $$

Commented by mathdave last updated on 07/Sep/20

has u have the right to stay in d forum dat is how i have the full right to stay on this forum bcos nobody is subscribing for anybody

$${has}\:{u}\:{have}\:{the}\:{right}\:{to}\:{stay}\:{in}\:{d}\:{forum} \\ $$$${dat}\:{is}\:{how}\:{i}\:{have}\:{the}\:{full}\:{right}\:{to}\:{stay} \\ $$$${on}\:{this}\:{forum}\:{bcos}\:{nobody}\:{is} \\ $$$${subscribing}\:{for}\:{anybody}\: \\ $$

Commented by mathdave last updated on 07/Sep/20

sorry mr mjs d question was a typo,i already corrected it

$${sorry}\:{mr}\:{mjs}\:{d}\:{question}\:{was}\:{a}\:{typo},{i} \\ $$$${already}\:{corrected}\:{it}\: \\ $$

Commented by mathdave last updated on 07/Sep/20

if u dont like u re d one that will leave for me in ds group

$${if}\:{u}\:{dont}\:{like}\:{u}\:{re}\:{d}\:{one}\:{that}\:{will}\:{leave} \\ $$$${for}\:{me}\:{in}\:{ds}\:{group} \\ $$

Commented by ajfour last updated on 07/Sep/20

stay in the forum, with some decent conversation and words, i have been on the forum for some yesrs; never witnessed anyone using abusive words..kindly you too, abide by the same decorum..

$${stay}\:{in}\:{the}\:{forum},\:{with}\:{some}\:{decent} \\ $$$${conversation}\:{and}\:{words},\:{i}\:{have}\:{been} \\ $$$${on}\:{the}\:{forum}\:{for}\:{some}\:{yesrs};\:{never} \\ $$$${witnessed}\:{anyone}\:{using}\:{abusive} \\ $$$${words}..{kindly}\:{you}\:{too},\:{abide}\:{by}\:{the} \\ $$$${same}\:{decorum}.. \\ $$

Commented by MJS_new last updated on 07/Sep/20

The question wasn′t a typo, you have done this before. You want to make others look like fools but most of us here have become friends over the years... no more to tell you, I′ll ignore you from now on.

$$\mathrm{The}\:\mathrm{question}\:\mathrm{wasn}'\mathrm{t}\:\mathrm{a}\:\mathrm{typo},\:\mathrm{you}\:\mathrm{have}\:\mathrm{done} \\ $$$$\mathrm{this}\:\mathrm{before}.\:\mathrm{You}\:\mathrm{want}\:\mathrm{to}\:\mathrm{make}\:\mathrm{others}\:\mathrm{look} \\ $$$$\mathrm{like}\:\mathrm{fools}\:\mathrm{but}\:\mathrm{most}\:\mathrm{of}\:\mathrm{us}\:\mathrm{here}\:\mathrm{have}\:\mathrm{become} \\ $$$$\mathrm{friends}\:\mathrm{over}\:\mathrm{the}\:\mathrm{years}...\:\mathrm{no}\:\mathrm{more}\:\mathrm{to}\:\mathrm{tell}\:\mathrm{you}, \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{ignore}\:\mathrm{you}\:\mathrm{from}\:\mathrm{now}\:\mathrm{on}. \\ $$

Commented by mathmax by abdo last updated on 07/Sep/20

how are sir mjs ?

$$\mathrm{how}\:\mathrm{are}\:\mathrm{sir}\:\mathrm{mjs}\:? \\ $$

Commented by Tawa11 last updated on 06/Sep/21

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

Answered by mathmax by abdo last updated on 07/Sep/20

I =∫_0 ^1 (√(1+x^2 ))dx by parts I =[x(√(1+x^2 ))]_0 ^1 −∫_0 ^1 x ((2x)/(2(√(1+x^2 )))) dx =(√2)−∫_0 ^1 (x^2 /(√(1+x^2 ))) dx =(√2)−∫_0 ^1 ((1+x^2 −1)/(√(1+x^2 ))) dx =(√2)−∫_0 ^1 (√(1+x^2 )) +∫_0 ^1 (dx/(√(1+x^2 ))) ⇒2I =(√2)+[ln(x+(√(1+x^2 ))]_0 ^1 =(√2)+ ln(1+(√2)) ⇒ I =((√2)/2) +(1/2)ln(1+(√2))

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts}\:\mathrm{I}\:=\left[\mathrm{x}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}\:\:\frac{\mathrm{2x}}{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx} \\ $$$$=\sqrt{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx}\:\:=\sqrt{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx} \\ $$$$=\sqrt{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\:\Rightarrow\mathrm{2I}\:=\sqrt{\mathrm{2}}+\left[\mathrm{ln}\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} \right. \\ $$$$=\sqrt{\mathrm{2}}+\:\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:\Rightarrow\:\mathrm{I}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$

Answered by 1549442205PVT last updated on 10/Sep/20

Put (1/x^2 )+1=t^2 ⇒−((2dx)/x^3 )=2tdt ⇒dx=−x^3 tdt,(√(1+x^2 ))=xt.Hence, F=∫_0 ^1 (√(1+x^2 ))dx=∫_(√2) ^∞ x^4 t^2 dt=∫_(√2) ^∞ ((t^2 dt)/((t^2 −1)^2 )) =∫_(√2) ^∞ (((t^2 −1)+1)/((t^2 −1)^2 ))dt=∫_(√2) ^∞ (dt/(t^2 −1))+∫_(√2) ^∞ (dt/((t^2 −1)^2 )) =∫_(√2) ^∞ (A+B) A=(1/(t^2 −1))=(1/2)((1/(t−1))−(1/(t+1))).Put(1/(t−1))=a,(1/(t+1))=b ⇒A=ab=(1/2)(a−b) we get B=A^2 =[(1/2)(a−b)]^2 =(1/4)(a^2 +b^2 ) −(1/2)ab=(1/4)(a^2 +b^2 )−(1/4)(a−b).Hence ⇒A+B=(1/4)(a^2 +b^2 )+(1/4)(a−b)=(1/4)[(1/((t−1)^2 ))+(1/((t+1)^2 ))]+(1/4)((1/(t−1))−(1/(t+1))) F=∫_(√2) ^∞ (A+B)=(1/4)∫_(√2) ^∞ (dt/((t−1)^2 ))+(1/4)∫_(√2) ^∞ (dt/((t+1)^2 )) +(1/4)∫_(√2) ^∞ ((1/(t−1))−(1/(t+1)))dt =[−(1/(4(t−1)))−(1/(4(t+1)))]_(√2) ^∞ +(1/4)ln∣((t−1)/(t+1))∣_(√2) ^∞ =−(1/(4((√2)−1)))−(1/(4((√2)+1)))−(1/4)ln(((√2)−1)/( (√2)+1)) =(1/4)((√2)+1+(√2)−1)−(1/4)ln((1/(((√2)+1)^2 ))) =((√2)/2)+(1/2)ln((√2)+1)

$$\mathrm{Put}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\mathrm{1}=\mathrm{t}^{\mathrm{2}} \Rightarrow−\frac{\mathrm{2dx}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{2tdt} \\ $$$$\Rightarrow\mathrm{dx}=−\mathrm{x}^{\mathrm{3}} \mathrm{tdt},\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\mathrm{xt}.\mathrm{Hence}, \\ $$$$\mathrm{F}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\sqrt{\mathrm{2}}} ^{\infty} \mathrm{x}^{\mathrm{4}} \mathrm{t}^{\mathrm{2}} \mathrm{dt}=\int_{\sqrt{\mathrm{2}}} ^{\infty} \frac{\mathrm{t}^{\mathrm{2}} \mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\int_{\sqrt{\mathrm{2}}} ^{\infty} \frac{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt}=\int_{\sqrt{\mathrm{2}}} ^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}+\int_{\sqrt{\mathrm{2}}} ^{\infty} \frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\int_{\sqrt{\mathrm{2}}} ^{\infty} \left(\mathrm{A}+\mathrm{B}\right) \\ $$$$\mathrm{A}=\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}\right).\mathrm{Put}\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}=\mathrm{a},\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}=\mathrm{b} \\ $$$$\Rightarrow\mathrm{A}=\mathrm{ab}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}−\mathrm{b}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{B}=\mathrm{A}^{\mathrm{2}} =\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}−\mathrm{b}\right)\right]^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ab}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{a}−\mathrm{b}\right).\mathrm{Hence} \\ $$$$\Rightarrow\mathrm{A}+\mathrm{B}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{a}−\mathrm{b}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{1}}{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} }\right]+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}\right) \\ $$$$\mathrm{F}=\int_{\sqrt{\mathrm{2}}} ^{\infty} \left(\mathrm{A}+\mathrm{B}\right)=\frac{\mathrm{1}}{\mathrm{4}}\int_{\sqrt{\mathrm{2}}} ^{\infty} \frac{\mathrm{dt}}{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\int_{\sqrt{\mathrm{2}}} ^{\infty} \frac{\mathrm{dt}}{\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$+\frac{\mathrm{1}}{\mathrm{4}}\int_{\sqrt{\mathrm{2}}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}\right)\mathrm{dt} \\ $$$$=\left[−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{t}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{t}+\mathrm{1}\right)}\right]_{\sqrt{\mathrm{2}}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\frac{\mathrm{t}−\mathrm{1}}{\mathrm{t}+\mathrm{1}}\mid_{\sqrt{\mathrm{2}}} ^{\infty} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\sqrt{\mathrm{2}}+\mathrm{1}+\sqrt{\mathrm{2}}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$

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