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Question Number 165328 by mnjuly1970 last updated on 30/Jan/22

       prove that             Nice   Integral         𝛗=∫_0 ^( 1) (( tan^( βˆ’1)  (x^( (3/2)) ))/x^( 2) ) dx  =((Ο€ + (√3) ln(7 +4(√3) ))/4)             β–   m.n        βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’

$$ \\ $$$$\:\:\:\:\:{prove}\:{that} \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:\:\mathscr{N}{ice}\:\:\:\mathscr{I}{ntegral} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{tan}^{\:βˆ’\mathrm{1}} \:\left({x}^{\:\frac{\mathrm{3}}{\mathrm{2}}} \right)}{{x}^{\:\mathrm{2}} }\:{dx}\:\:=\frac{\pi\:+\:\sqrt{\mathrm{3}}\:{ln}\left(\mathrm{7}\:+\mathrm{4}\sqrt{\mathrm{3}}\:\right)}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:\:{m}.{n} \\ $$$$\:\:\:\:\:\:βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’\:\:\: \\ $$

Answered by Lordose last updated on 30/Jan/22

Ξ¦ = ∫_0 ^( 1) ((tan^(βˆ’1) (x^(3/2) ))/x^2 )dx =^(IBP) βˆ’((tan^(βˆ’1) (x^(3/2) ))/x)∣_0 ^1 +(3/2)∫_0 ^( 1) (x^((1/2)βˆ’1) /(1+x^3 ))dx  Ξ¦ = βˆ’(𝛑/4) + (3/2)∫_0 ^( 1) (1/( (√x)(1+x^3 )))dx  Ξ¦ = βˆ’(𝛑/4) + 3∫_0 ^( 1) (1/((1+x^6 )))dx  Ξ¦ = βˆ’(𝛑/4) + 3Ξ£_(n=1) ^∞ (βˆ’1)^(nβˆ’1) ∫_0 ^( 1) x^(6nβˆ’6) dx  Ξ¦ = βˆ’(𝛑/4) + 3Ξ£_(n= 1) ^∞ (((βˆ’1)^(nβˆ’1) )/(6nβˆ’5))  Ξ¦ = βˆ’(𝛑/4) + 3((𝛑/6) + ((coth^(βˆ’1) ((√3)))/( (√3))))  Ξ¦ = (𝛑/4) + (√3)coth^(βˆ’1) ((√3))

$$\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{tan}^{βˆ’\mathrm{1}} \left(\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\overset{\boldsymbol{\mathrm{IBP}}} {=}βˆ’\frac{\mathrm{tan}^{βˆ’\mathrm{1}} \left(\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)}{\mathrm{x}}\mid_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}βˆ’\mathrm{1}} }{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }\mathrm{dx} \\ $$$$\Phi\:=\:βˆ’\frac{\boldsymbol{\pi}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}\left(\mathrm{1}+\mathrm{x}^{\mathrm{3}} \right)}\mathrm{dx} \\ $$$$\Phi\:=\:βˆ’\frac{\boldsymbol{\pi}}{\mathrm{4}}\:+\:\mathrm{3}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{6}} \right)}\mathrm{dx} \\ $$$$\Phi\:=\:βˆ’\frac{\boldsymbol{\pi}}{\mathrm{4}}\:+\:\mathrm{3}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(βˆ’\mathrm{1}\right)^{\mathrm{n}βˆ’\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{x}^{\mathrm{6n}βˆ’\mathrm{6}} \mathrm{dx} \\ $$$$\Phi\:=\:βˆ’\frac{\boldsymbol{\pi}}{\mathrm{4}}\:+\:\mathrm{3}\underset{\mathrm{n}=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{\mathrm{n}βˆ’\mathrm{1}} }{\mathrm{6n}βˆ’\mathrm{5}} \\ $$$$\Phi\:=\:βˆ’\frac{\boldsymbol{\pi}}{\mathrm{4}}\:+\:\mathrm{3}\left(\frac{\boldsymbol{\pi}}{\mathrm{6}}\:+\:\frac{\mathrm{coth}^{βˆ’\mathrm{1}} \left(\sqrt{\mathrm{3}}\right)}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\Phi\:=\:\frac{\boldsymbol{\pi}}{\mathrm{4}}\:+\:\sqrt{\mathrm{3}}\mathrm{coth}^{βˆ’\mathrm{1}} \left(\sqrt{\mathrm{3}}\right) \\ $$

Answered by mnjuly1970 last updated on 30/Jan/22

          βˆ’βˆ’βˆ’ solutionβˆ’βˆ’βˆ’        𝛗=^(i.b.p)  [((βˆ’1)/x) tan^( βˆ’1)  ( x^( (3/2)) )]_0 ^1 + (3/2) ∫_0 ^( 1)  (1/( (√x) .( 1 + x^( 3) ))) dx             = βˆ’(Ο€/4)  + (3/2) Ξ©    where  Ξ© = ∫_0 ^( 1) (( dx)/( (√x) (1 +x^( 3) )))      Ξ© =^((√x) =t)   ∫_0 ^( 1) (( 2)/(1 + x^( 6) )) dx =∫_0 ^( 1) (( x^( 4) +1 βˆ’ (x^( 4) βˆ’1 ))/(1+ x^( 6) )) dx           = ∫_0 ^( 1) (( (x^( 4) βˆ’x^( 2) +1) + x^( 2) )/(1 + x^( 6) ))dx + ∫_0 ^( 1) ((1βˆ’x^( 2) )/(1 βˆ’x^( 2) + x^( 4) )) dx        =   (Ο€/4) + ∫_0 ^( 1) (( 3x^( 2) )/( 1+ (x^( 3) )^( 2) )) dx βˆ’βˆ«_0 ^( 1) ((1βˆ’ x^( βˆ’2) )/(( x + x^( βˆ’1) )^( 2) βˆ’3))dx       = (Ο€/4)  + (Ο€/(   4  )) + ∫_2 ^( ∞) (( dx)/(x^( 2) βˆ’3))        = (Ο€/2)  + ∫_2 ^( ∞)  (dx/(( x βˆ’(√3) )( x +(√3) )))        = (Ο€/2)  + (1/(2(√3))) { [ln(((xβˆ’(√3))/(x+(√3))) )]_2 ^∞ }          = (Ο€/2) + (1/(2(√3))) ln(((2+(√3))/(2βˆ’(√3))) )           = (Ο€/2) + (1/(2(√3))) ln (7 +4 (√3) )         ∴    𝛗 = (Ο€/4) + ((√3)/4) ln ( 7 + 4 (√3) )   β–  m.n

$$\:\:\:\:\:\:\:\:\:\:βˆ’βˆ’βˆ’\:{solution}βˆ’βˆ’βˆ’ \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\overset{{i}.{b}.{p}} {=}\:\left[\frac{βˆ’\mathrm{1}}{{x}}\:{tan}^{\:βˆ’\mathrm{1}} \:\left(\:{x}^{\:\frac{\mathrm{3}}{\mathrm{2}}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\:\frac{\mathrm{3}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{x}}\:.\left(\:\mathrm{1}\:+\:{x}^{\:\mathrm{3}} \right)}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:βˆ’\frac{\pi}{\mathrm{4}}\:\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:\Omega\:\:\:\:{where}\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{dx}}{\:\sqrt{{x}}\:\left(\mathrm{1}\:+{x}^{\:\mathrm{3}} \right)} \\ $$$$\:\:\:\:\Omega\:\overset{\sqrt{{x}}\:={t}} {=}\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{2}}{\mathrm{1}\:+\:{x}^{\:\mathrm{6}} }\:{dx}\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}^{\:\mathrm{4}} +\mathrm{1}\:βˆ’\:\left({x}^{\:\mathrm{4}} βˆ’\mathrm{1}\:\right)}{\mathrm{1}+\:{x}^{\:\mathrm{6}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\left({x}^{\:\mathrm{4}} βˆ’{x}^{\:\mathrm{2}} +\mathrm{1}\right)\:+\:{x}^{\:\mathrm{2}} }{\mathrm{1}\:+\:{x}^{\:\mathrm{6}} }{dx}\:+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}βˆ’{x}^{\:\mathrm{2}} }{\mathrm{1}\:βˆ’{x}^{\:\mathrm{2}} +\:{x}^{\:\mathrm{4}} }\:{dx} \\ $$$$\:\:\:\:\:\:=\:\:\:\frac{\pi}{\mathrm{4}}\:+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{3}{x}^{\:\mathrm{2}} }{\:\mathrm{1}+\:\left({x}^{\:\mathrm{3}} \right)^{\:\mathrm{2}} }\:{dx}\:βˆ’\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}βˆ’\:{x}^{\:βˆ’\mathrm{2}} }{\left(\:{x}\:+\:{x}^{\:βˆ’\mathrm{1}} \right)^{\:\mathrm{2}} βˆ’\mathrm{3}}{dx} \\ $$$$\:\:\:\:\:=\:\frac{\pi}{\mathrm{4}}\:\:+\:\frac{\pi}{\:\:\:\mathrm{4}\:\:}\:+\:\int_{\mathrm{2}} ^{\:\infty} \frac{\:{dx}}{{x}^{\:\mathrm{2}} βˆ’\mathrm{3}} \\ $$$$\:\:\:\:\:\:=\:\frac{\pi}{\mathrm{2}}\:\:+\:\int_{\mathrm{2}} ^{\:\infty} \:\frac{{dx}}{\left(\:{x}\:βˆ’\sqrt{\mathrm{3}}\:\right)\left(\:{x}\:+\sqrt{\mathrm{3}}\:\right)} \\ $$$$\:\:\:\:\:\:=\:\frac{\pi}{\mathrm{2}}\:\:+\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\left\{\:\left[{ln}\left(\frac{{x}βˆ’\sqrt{\mathrm{3}}}{{x}+\sqrt{\mathrm{3}}}\:\right)\right]_{\mathrm{2}} ^{\infty} \right\} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\pi}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}βˆ’\sqrt{\mathrm{3}}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{\pi}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:{ln}\:\left(\mathrm{7}\:+\mathrm{4}\:\sqrt{\mathrm{3}}\:\right) \\ $$$$\:\:\:\:\:\:\:\therefore\:\:\:\:\boldsymbol{\phi}\:=\:\frac{\pi}{\mathrm{4}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:{ln}\:\left(\:\mathrm{7}\:+\:\mathrm{4}\:\sqrt{\mathrm{3}}\:\right)\:\:\:\blacksquare\:{m}.{n}\:\:\:\:\:\: \\ $$$$\:\: \\ $$

Answered by Ar Brandon last updated on 30/Jan/22

𝛗=∫_0 ^1 ((tan^(βˆ’1) (x^(3/2) ))/x^2 )dx=βˆ’[(1/x)tan^(βˆ’1) (x^(3/2) )]_0 ^1 +(3/2)∫_0 ^1 (x^(βˆ’(1/2)) /(1+x^3 ))dx      =βˆ’(Ο€/4)+3∫_0 ^1 (dt/(1+t^6 ))=βˆ’(Ο€/4)+3∫_0 ^1 (dt/((t^2 +1)(t^4 βˆ’t^2 +1)))      =βˆ’(Ο€/4)+∫_0 ^1 ((1/(t^2 +1))βˆ’((t^2 βˆ’2)/(t^4 βˆ’t^2 +1)))dt=βˆ’(Ο€/4)+(Ο€/4)βˆ’βˆ«_0 ^1 ((t^2 βˆ’2)/(t^4 βˆ’t^2 +1))dt      =βˆ’(1/2)∫_0 ^1 ((3(t^2 βˆ’1)βˆ’(t^2 +1))/(t^4 βˆ’t^2 +1))dt=(1/2)∫_0 ^1 ((t^2 +1)/(t^4 βˆ’t^2 +1))dtβˆ’(3/2)∫_0 ^1 ((t^2 βˆ’1)/(t^4 βˆ’t^2 +1))dt      =(1/2)∫_0 ^1 ((1+(1/t^2 ))/((tβˆ’(1/t))^2 +1))dtβˆ’(3/2)∫_0 ^1 ((1βˆ’(1/t^2 ))/((t+(1/t))^2 βˆ’3))dt      =(1/2)[arctan(((t^2 βˆ’1)/t))]_0 ^1 +((√3)/4)[ln∣((t^2 +(√3)t+1)/(t^2 βˆ’(√3)t+1))∣]_0 ^1       =(1/2)((Ο€/2))+((√3)/4)ln∣((2+(√3))/(2βˆ’(√3)))∣=(Ο€/4)+((√3)/4)ln(2+(√3))^2       =(Ο€/4)+((√3)/4)ln(7+4(√3))

$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{βˆ’\mathrm{1}} \left({x}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)}{{x}^{\mathrm{2}} }{dx}=βˆ’\left[\frac{\mathrm{1}}{{x}}\mathrm{tan}^{βˆ’\mathrm{1}} \left({x}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{βˆ’\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{x}^{\mathrm{3}} }{dx} \\ $$$$\:\:\:\:=βˆ’\frac{\pi}{\mathrm{4}}+\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{6}} }=βˆ’\frac{\pi}{\mathrm{4}}+\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{4}} βˆ’{t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:\:\:=βˆ’\frac{\pi}{\mathrm{4}}+\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}βˆ’\frac{{t}^{\mathrm{2}} βˆ’\mathrm{2}}{{t}^{\mathrm{4}} βˆ’{t}^{\mathrm{2}} +\mathrm{1}}\right){dt}=βˆ’\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} βˆ’\mathrm{2}}{{t}^{\mathrm{4}} βˆ’{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\:\:\:\:=βˆ’\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{3}\left({t}^{\mathrm{2}} βˆ’\mathrm{1}\right)βˆ’\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{{t}^{\mathrm{4}} βˆ’{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} βˆ’{t}^{\mathrm{2}} +\mathrm{1}}{dt}βˆ’\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} βˆ’\mathrm{1}}{{t}^{\mathrm{4}} βˆ’{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}βˆ’\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{1}}{dt}βˆ’\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} βˆ’\mathrm{3}}{dt} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{arctan}\left(\frac{{t}^{\mathrm{2}} βˆ’\mathrm{1}}{{t}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left[\mathrm{ln}\mid\frac{{t}^{\mathrm{2}} +\sqrt{\mathrm{3}}{t}+\mathrm{1}}{{t}^{\mathrm{2}} βˆ’\sqrt{\mathrm{3}}{t}+\mathrm{1}}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{ln}\mid\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}βˆ’\sqrt{\mathrm{3}}}\mid=\frac{\pi}{\mathrm{4}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\pi}{\mathrm{4}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{ln}\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right) \\ $$

Commented by mnjuly1970 last updated on 31/Jan/22

thanks alot sir brandon

$${thanks}\:{alot}\:{sir}\:{brandon} \\ $$$$ \\ $$

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