Question Number 124644 by benjo_mathlover last updated on 05/Dec/20
$$\underset{\mathrm{2}/\sqrt{\mathrm{3}}} {\overset{\mathrm{2}} {\int}}\:\frac{\mathrm{cos}\:\left(\mathrm{sec}^{−\mathrm{1}} {x}\right)}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:{dx}\: \\ $$$$\underset{\:\sqrt{\mathrm{2}}} {\overset{\mathrm{2}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} \left(\mathrm{sec}^{−\mathrm{1}} {x}\right)}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:{dx}\: \\ $$
Answered by liberty last updated on 05/Dec/20
![(1) ∫ _(2/( (√3))) ^( 2) ((cos (sec^(−1) x))/(x(√(x^2 −1)))) dx put sec^(−1) x = j , x = sec j and dx = sec j tan j dj where upper limit j=(π/3) and lower limit j=(π/6) ∫ _(π/6) ^( π/3) ((cos j)/(sec j tan j)) (sec j tan j )dj I= [ sin j ]_(π/6) ^(π/3) = ((√3)/2) − (1/2) = (1/2)((√3)−1)](https://www.tinkutara.com/question/Q124648.png)
$$\left(\mathrm{1}\right)\:\int\underset{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}} {\overset{\:\mathrm{2}} {\:}}\frac{\mathrm{cos}\:\left(\mathrm{sec}^{−\mathrm{1}} {x}\right)}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:{dx}\: \\ $$$$\:{put}\:\mathrm{sec}^{−\mathrm{1}} {x}\:=\:{j}\:,\:{x}\:=\:\mathrm{sec}\:{j}\:{and}\:{dx}\:=\:\mathrm{sec}\:{j}\:\mathrm{tan}\:{j}\:{dj} \\ $$$${where}\:{upper}\:{limit}\:{j}=\frac{\pi}{\mathrm{3}}\:{and}\:{lower}\:{limit}\:{j}=\frac{\pi}{\mathrm{6}} \\ $$$$\int\overset{\:\pi/\mathrm{3}} {\:}_{\pi/\mathrm{6}} \frac{\mathrm{cos}\:{j}}{\mathrm{sec}\:{j}\:\mathrm{tan}\:{j}}\:\left(\mathrm{sec}\:{j}\:\mathrm{tan}\:{j}\:\right){dj} \\ $$$${I}=\:\left[\:\mathrm{sin}\:{j}\:\right]_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$
Answered by Dwaipayan Shikari last updated on 05/Dec/20
![∫_(√2) ^2 ((sec^2 (sec^(−1) x))/(x(√(x^2 −1))))dx sec^(−1) x =t⇒(1/(x(√(x^2 −1))))=(dt/dx) =∫_(π/4) ^(π/3) sec^2 (t)dt = [tan(t)]_(π/4) ^(π/3) = (√3)−1](https://www.tinkutara.com/question/Q124649.png)
$$\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \frac{{sec}^{\mathrm{2}} \left({sec}^{−\mathrm{1}} {x}\right)}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx}\:\:\:\:\:\:\:\:{sec}^{−\mathrm{1}} {x}\:={t}\Rightarrow\frac{\mathrm{1}}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\frac{{dt}}{{dx}} \\ $$$$=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} {sec}^{\mathrm{2}} \left({t}\right){dt}\:=\:\left[{tan}\left({t}\right)\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} =\:\sqrt{\mathrm{3}}−\mathrm{1} \\ $$