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Question Number 112651 by bemath last updated on 09/Sep/20

(1)lim_(x→a) (x−a) cosec (((πx)/a)) ? (2) ∫ (√(x^2 +1)) dx , by using Euler′s substitution

$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\left(\mathrm{x}−\mathrm{a}\right)\:\mathrm{cosec}\:\left(\frac{\pi\mathrm{x}}{\mathrm{a}}\right)\:? \\ $$$$\left(\mathrm{2}\right)\:\int\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{dx}\:,\:\mathrm{by}\:\mathrm{using}\:\mathrm{Euler}'\mathrm{s} \\ $$$$\mathrm{substitution} \\ $$

Answered by john santu last updated on 09/Sep/20

(★) by Euler′s substitution let (√(x^2 +1)) = x+q ⇒x^2 +1 = x^2 +2qx+q^2 2qx+q^2 = 1 ⇒x = ((1−q^2 )/(2q)) dx = ((−4q^2 −2(1−q^2 ))/(4q^2 )) dq dx = ((−2q^2 −2)/(4q^2 )) dq = ((−q^2 −1)/(2q^2 )) dq so I = ∫ (√(x^2 +1)) dx I= ∫ (((1−q^2 )/(2q))+q).(((−q^2 −1)/(2q^2 )))dq I= −∫ (((q^2 +1)/(2q)))(((q^2 +1)/(2q^2 )))dq I=−(1/4)∫ (((q^4 +2q^2 +1)/q^3 ))dq I=−(1/4)∫ (q+(2/q)+q^(−3) )dq I=−(1/4)((1/2)q^2 +2ln (q)−(1/(2q^2 )))+c I=−(1/8)((√(x^2 +1))−x)^2 −(1/2)ln ∣(√(x^2 +1))−x∣+(1/(8((√(x^2 +1))−x)^2 )) + c

$$\left(\bigstar\right)\:{by}\:{Euler}'{s}\:{substitution}\: \\ $$$${let}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:=\:{x}+{q}\:\Rightarrow{x}^{\mathrm{2}} +\mathrm{1}\:=\:{x}^{\mathrm{2}} +\mathrm{2}{qx}+{q}^{\mathrm{2}} \\ $$$$\:\:\mathrm{2}{qx}+{q}^{\mathrm{2}} \:=\:\mathrm{1}\:\Rightarrow{x}\:=\:\frac{\mathrm{1}−{q}^{\mathrm{2}} }{\mathrm{2}{q}} \\ $$$$\:\:{dx}\:=\:\frac{−\mathrm{4}{q}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−{q}^{\mathrm{2}} \right)}{\mathrm{4}{q}^{\mathrm{2}} }\:{dq} \\ $$$$\:\:{dx}\:=\:\frac{−\mathrm{2}{q}^{\mathrm{2}} −\mathrm{2}}{\mathrm{4}{q}^{\mathrm{2}} }\:{dq}\:=\:\frac{−{q}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{q}^{\mathrm{2}} }\:{dq} \\ $$$${so}\:{I}\:=\:\int\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\: \\ $$$${I}=\:\int\:\left(\frac{\mathrm{1}−{q}^{\mathrm{2}} }{\mathrm{2}{q}}+{q}\right).\left(\frac{−{q}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{q}^{\mathrm{2}} }\right){dq} \\ $$$${I}=\:−\int\:\left(\frac{{q}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{q}}\right)\left(\frac{{q}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{q}^{\mathrm{2}} }\right){dq} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{4}}\int\:\left(\frac{{q}^{\mathrm{4}} +\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}}{{q}^{\mathrm{3}} }\right){dq} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{4}}\int\:\left({q}+\frac{\mathrm{2}}{{q}}+{q}^{−\mathrm{3}} \right){dq} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}{q}^{\mathrm{2}} +\mathrm{2ln}\:\left({q}\right)−\frac{\mathrm{1}}{\mathrm{2}{q}^{\mathrm{2}} }\right)+{c} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{8}}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−{x}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−{x}\mid+\frac{\mathrm{1}}{\mathrm{8}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−{x}\right)^{\mathrm{2}} }\:+\:{c} \\ $$

Answered by bobhans last updated on 09/Sep/20

(⧫) lim_(x→a) (x−a) cosec (((πx)/a)) = L L= lim_(x→a) (((x−a))/(sin (((πx)/a))))= lim_(x→a) (1/((π/a)cos (((πx)/a)))) L = (a/π) lim_(x→a) (1/(cos (((πx)/a)))) = −(a/π)

$$\left(\blacklozenge\right)\:\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\left(\mathrm{x}−\mathrm{a}\right)\:\mathrm{cosec}\:\left(\frac{\pi\mathrm{x}}{\mathrm{a}}\right)\:=\:\mathrm{L} \\ $$$$\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\mathrm{a}\right)}{\mathrm{sin}\:\left(\frac{\pi\mathrm{x}}{\mathrm{a}}\right)}=\:\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\frac{\pi}{\mathrm{a}}\mathrm{cos}\:\left(\frac{\pi\mathrm{x}}{\mathrm{a}}\right)} \\ $$$$\:\mathrm{L}\:=\:\frac{\mathrm{a}}{\pi}\:\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{cos}\:\left(\frac{\pi\mathrm{x}}{\mathrm{a}}\right)}\:=\:−\frac{\mathrm{a}}{\pi} \\ $$

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

2)Find ∫(√(x^2 +1)) dx Put x=tant⇒dx=(1+tan^2 t)dt.Then F=∫(√(x^2 +1)) dx=∫(√(1+tan^2 t)) (1+tan^2 t)dt F=∫(dt/(cos^3 t))dt=(1/2)∫(1/(sint))×((2sintcost)/(cos^4 t))dt =(1/2)∫(1/(sint))d((1/(cos^2 t)))=(1/2)×(1/(sintcos^2 t)) −(1/2)∫(1/(cos^2 t)).((1/(sint)))′dt=(1/2)×(1/(sintcos^2 t)) −(1/2)∫(1/(cos^2 t)).(−(1/(sin^2 t)))×costdt =(1/2)×(1/(sintcos^2 t))+(1/2)∫((d(sint))/(cos^2 tsin^2 t)) Put sint=u we get (1/2)∫((d(sint))/(cos^2 tsin^2 t))=(1/2)∫(du/(u^2 (1−u^2 ))) =(1/2)∫((1/u^2 )+(1/(1−u^2 )))du=−(1/(2u))−(1/2)∫(du/(u^2 −1)) =((−1)/(2u))−(1/2)×(1/2)ln∣((u−1)/(u+1))∣.Therefore, F=(1/(2sintcos^2 t))−(1/(2sint))−(1/4)ln∣((sint−1)/(sint+1))∣+C x=tant⇔cos^2 t=(1/(1+x^2 )),sint =((∣x∣)/( (√(1+x^2 )))) F=((∣x∣(√(1+x^2 )))/2)−(1/4)ln∣((∣x∣−(√(1+x^2 )))/(∣x∣+(√( 1+x^2 ))))∣+C Also see question 112313

$$\left.\mathrm{2}\right)\boldsymbol{\mathrm{Find}}\:\int\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}}\:\boldsymbol{\mathrm{dx}}\: \\ $$$$\mathrm{Put}\:\mathrm{x}=\mathrm{tant}\Rightarrow\mathrm{dx}=\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}\right)\mathrm{dt}.\mathrm{Then} \\ $$$$\mathrm{F}=\int\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{dx}=\int\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}}\:\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}\right)\mathrm{dt} \\ $$$$\mathrm{F}=\int\frac{\mathrm{dt}}{\mathrm{cos}^{\mathrm{3}} \mathrm{t}}\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\mathrm{sint}}×\frac{\mathrm{2sintcost}}{\mathrm{cos}^{\mathrm{4}} \mathrm{t}}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\mathrm{sint}}\mathrm{d}\left(\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{t}}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{sintcos}^{\mathrm{2}} \mathrm{t}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{t}}.\left(\frac{\mathrm{1}}{\mathrm{sint}}\right)'\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{sintcos}^{\mathrm{2}} \mathrm{t}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{t}}.\left(−\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}\right)×\mathrm{costdt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{sintcos}^{\mathrm{2}} \mathrm{t}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{d}\left(\mathrm{sint}\right)}{\mathrm{cos}^{\mathrm{2}} \mathrm{tsin}^{\mathrm{2}} \mathrm{t}} \\ $$$$\mathrm{Put}\:\mathrm{sint}=\mathrm{u}\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{d}\left(\mathrm{sint}\right)}{\mathrm{cos}^{\mathrm{2}} \mathrm{tsin}^{\mathrm{2}} \mathrm{t}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{u}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }\right)\mathrm{du}=−\frac{\mathrm{1}}{\mathrm{2u}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2u}}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{\mathrm{u}−\mathrm{1}}{\mathrm{u}+\mathrm{1}}\mid.\mathrm{Therefore}, \\ $$$$\boldsymbol{\mathrm{F}}=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\mathrm{sintcos}}^{\mathrm{2}} \boldsymbol{\mathrm{t}}}−\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\mathrm{sint}}}−\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{\mathrm{ln}}\mid\frac{\boldsymbol{\mathrm{sint}}−\mathrm{1}}{\boldsymbol{\mathrm{sint}}+\mathrm{1}}\mid+\boldsymbol{\mathrm{C}} \\ $$$$\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{tant}}\Leftrightarrow\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \mathrm{t}=\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} },\boldsymbol{\mathrm{sint}}\:=\frac{\mid\boldsymbol{\mathrm{x}}\mid}{\:\sqrt{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }} \\ $$$$\boldsymbol{\mathrm{F}}=\frac{\mid\mathrm{x}\mid\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{\mathrm{ln}}\mid\frac{\mid\boldsymbol{\mathrm{x}}\mid−\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mid\boldsymbol{\mathrm{x}}\mid+\sqrt{\:\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\mid+\boldsymbol{\mathrm{C}} \\ $$$$\mathrm{Also}\:\mathrm{see}\:\mathrm{question}\:\mathrm{112313} \\ $$

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