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Question Number 54616 by cesar.marval.larez@gmail.com last updated on 07/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19

a=tanx  da=sec^2 xdx  ∫((a^2 da)/(√(2+1+a^2 )))  ∫((a^2 +3−3)/(√(a^2 +3)))da  ∫(√(a^2 +3)) da−3∫(da/(√(a^2 +3)))  now use formula  (a/2)(√(a^2 +3)) +(3/2)ln(a+(√(a^2 +3)) )−3ln(a+(√(a^2 +3)) )+c  ((tanx)/2)(√(tan^2 x+3)) −(3/2)ln(tanx+(√(tan^2 x+3)) )+c

$${a}={tanx}\:\:{da}={sec}^{\mathrm{2}} {xdx} \\ $$$$\int\frac{{a}^{\mathrm{2}} {da}}{\sqrt{\mathrm{2}+\mathrm{1}+{a}^{\mathrm{2}} }} \\ $$$$\int\frac{{a}^{\mathrm{2}} +\mathrm{3}−\mathrm{3}}{\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}}{da} \\ $$$$\int\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}\:{da}−\mathrm{3}\int\frac{{da}}{\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}} \\ $$$${now}\:{use}\:{formula} \\ $$$$\frac{{a}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}\:+\frac{\mathrm{3}}{\mathrm{2}}{ln}\left({a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}\:\right)−\mathrm{3}{ln}\left({a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}\:\right)+{c} \\ $$$$\frac{{tanx}}{\mathrm{2}}\sqrt{{tan}^{\mathrm{2}} {x}+\mathrm{3}}\:−\frac{\mathrm{3}}{\mathrm{2}}{ln}\left({tanx}+\sqrt{{tan}^{\mathrm{2}} {x}+\mathrm{3}}\:\right)+{c} \\ $$

Answered by MJS last updated on 08/Feb/19

let′s be cheeky  ∫((sec^2  x tan^2  x)/(√(2+sec^2  x)))dx=       [t=(√(2+sec^2  x)) → x=arccos((1/(√(t^2 −2)))); dx=((sec^2  x (√(2+sec^2  x)))/(tan x))dt]  =∫tan x dt=∫(√(t^2 −3))dt=       [u=((√3)/3)t → dt=(√3)du]  =3∫(√(u^2 −1))du=       [v=arcosh u → du=sinh v dv]  =3∫sinh^2  v dv=(3/2)(v+sinh v cosh v)=  =(3/2)(u(√(u^2 −1))−arcosh u)=  =(1/2)(t(√(t^2 −3))−3arcosh ((t(√3))/3))=  =(1/2)(√(sec^4  x +sec^2  x −2))−(3/2)arcosh (√((2+sec^2  x)/3)) +C

$$\mathrm{let}'\mathrm{s}\:\mathrm{be}\:\mathrm{cheeky} \\ $$$$\int\frac{\mathrm{sec}^{\mathrm{2}} \:{x}\:\mathrm{tan}^{\mathrm{2}} \:{x}}{\sqrt{\mathrm{2}+\mathrm{sec}^{\mathrm{2}} \:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{2}+\mathrm{sec}^{\mathrm{2}} \:{x}}\:\rightarrow\:{x}=\mathrm{arccos}\left(\frac{\mathrm{1}}{\sqrt{{t}^{\mathrm{2}} −\mathrm{2}}}\right);\:{dx}=\frac{\mathrm{sec}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{2}+\mathrm{sec}^{\mathrm{2}} \:{x}}}{\mathrm{tan}\:{x}}{dt}\right] \\ $$$$=\int\mathrm{tan}\:{x}\:{dt}=\int\sqrt{{t}^{\mathrm{2}} −\mathrm{3}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{t}\:\rightarrow\:{dt}=\sqrt{\mathrm{3}}{du}\right] \\ $$$$=\mathrm{3}\int\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}{du}= \\ $$$$\:\:\:\:\:\left[{v}=\mathrm{arcosh}\:{u}\:\rightarrow\:{du}=\mathrm{sinh}\:{v}\:{dv}\right] \\ $$$$=\mathrm{3}\int\mathrm{sinh}^{\mathrm{2}} \:{v}\:{dv}=\frac{\mathrm{3}}{\mathrm{2}}\left({v}+\mathrm{sinh}\:{v}\:\mathrm{cosh}\:{v}\right)= \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\left({u}\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}−\mathrm{arcosh}\:{u}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({t}\sqrt{{t}^{\mathrm{2}} −\mathrm{3}}−\mathrm{3arcosh}\:\frac{{t}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{sec}^{\mathrm{4}} \:{x}\:+\mathrm{sec}^{\mathrm{2}} \:{x}\:−\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{arcosh}\:\sqrt{\frac{\mathrm{2}+\mathrm{sec}^{\mathrm{2}} \:{x}}{\mathrm{3}}}\:+{C} \\ $$

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