Question Number 210082 by klipto last updated on 30/Jul/24
$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}−\boldsymbol{\mathrm{t}}+\boldsymbol{\mathrm{tx}}^{\mathrm{2}} \right)}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{1}}\mathrm{dx}=\left(\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \left(\sqrt{\boldsymbol{\mathrm{t}}}\right)\right)^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{anyone}}.. \\ $$
Answered by Frix last updated on 30/Jul/24
![Simply use Feynman′s Technique ∫_0 ^1 ((ln (tx^2 −t+1))/(x^2 −1))dx =f(t) f(0)=0 f′(t)= ∫_0 ^1 (∂/∂t)[((ln (tx^2 −t+1))/(x^2 −1))]dx= =∫_0 ^1 (dx/(tx^2 −t+1)) =^(u=((√t)/( (√(1−t))))x) =(1/( (√t)(√(1−t))))∫_0 ^((√t)/( (√(1−t)))) (du/(u^2 +1))= =(1/( (√t)(√(1−t))))[tan^(−1) u]_0 ^((√t)/( (√(1−t)))) =((tan^(−1) ((√t)/( (√(1−t)))))/( (√t)(√(1−t))))= =((sin^(−1) (√t))/( (√t)(√(1−t)))) f(t)=∫((sin^(−1) (√t))/( (√t)(√(1−αt))))dt =^(v=sin^(−1) (√t)) 2∫vdv=v^2 = =(sin^(−1) (√t))^2 +C f(0)=0 ⇒ C=0 f(t)=(sin^(−1) (√t))^2](https://www.tinkutara.com/question/Q210083.png)
$$\mathrm{Simply}\:\mathrm{use}\:\mathrm{Feynman}'\mathrm{s}\:\mathrm{Technique} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left({tx}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}\:={f}\left({t}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}'\left({t}\right)=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\partial}{\partial{t}}\left[\frac{\mathrm{ln}\:\left({tx}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\right]{dx}= \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{tx}^{\mathrm{2}} −{t}+\mathrm{1}}\:\overset{{u}=\frac{\sqrt{{t}}}{\:\sqrt{\mathrm{1}−{t}}}{x}} {=} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{t}}\sqrt{\mathrm{1}−{t}}}\underset{\mathrm{0}} {\overset{\frac{\sqrt{{t}}}{\:\sqrt{\mathrm{1}−{t}}}} {\int}}\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{t}}\sqrt{\mathrm{1}−{t}}}\left[\mathrm{tan}^{−\mathrm{1}} \:{u}\right]_{\mathrm{0}} ^{\frac{\sqrt{{t}}}{\:\sqrt{\mathrm{1}−{t}}}} =\frac{\mathrm{tan}^{−\mathrm{1}} \:\frac{\sqrt{{t}}}{\:\sqrt{\mathrm{1}−{t}}}}{\:\sqrt{{t}}\sqrt{\mathrm{1}−{t}}}= \\ $$$$=\frac{\mathrm{sin}^{−\mathrm{1}} \:\sqrt{{t}}}{\:\sqrt{{t}}\sqrt{\mathrm{1}−{t}}} \\ $$$${f}\left({t}\right)=\int\frac{\mathrm{sin}^{−\mathrm{1}} \:\sqrt{{t}}}{\:\sqrt{{t}}\sqrt{\mathrm{1}−\alpha{t}}}{dt}\:\overset{{v}=\mathrm{sin}^{−\mathrm{1}} \:\sqrt{{t}}} {=}\:\mathrm{2}\int{vdv}={v}^{\mathrm{2}} = \\ $$$$=\left(\mathrm{sin}^{−\mathrm{1}} \:\sqrt{{t}}\right)^{\mathrm{2}} +{C} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\:{C}=\mathrm{0} \\ $$$${f}\left({t}\right)=\left(\mathrm{sin}^{−\mathrm{1}} \:\sqrt{{t}}\right)^{\mathrm{2}} \\ $$
Commented by klipto last updated on 30/Jul/24
$$\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{bro}} \\ $$
Commented by klipto last updated on 30/Jul/24
$$\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{pdf}}\:\boldsymbol{\mathrm{do}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{channel}} \\ $$