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Question-70253




Question Number 70253 by oyemi kemewari last updated on 02/Oct/19
Commented by mathmax by abdo last updated on 02/Oct/19
let I =∫ u^2 (√(u^2 −2))du changement u=(√2)ch(x) give  I =∫ 2ch^2 (x)(√2)sh(x)(√2)sh(x)dx  =4 ∫  ch^2 x sh^2 x dx =4 ∫ (ch(x)sh(x))^2 dx  =4 ∫((1/2)sh(2x))^2 dx =∫  sh^2 (2x)dx =∫((ch(4x)−1)/2)dx  =−(x/2) +(1/8)sh(4x)dx  +C  =−(x/2) +(1/8){((e^(4x) −e^(−4x) )/2)} +C  =−(x/2) +(1/(16))(e^(4x) −e^(−4x) ) +C  but x=argch((u/( (√2))))  =ln((u/( (√2)))+(√((u^2 /2)−1))) ⇒  I =−(1/2)ln((u/( (√2)))+(√((u^2 /2)−1)))+(1/(16)){((u/( (√2)))+(√((u^2 /2)−1)))^4 −((u/( (√2)))+(√((u^2 /2)−1)))^(−4) }+C  =
$${let}\:{I}\:=\int\:{u}^{\mathrm{2}} \sqrt{{u}^{\mathrm{2}} −\mathrm{2}}{du}\:{changement}\:{u}=\sqrt{\mathrm{2}}{ch}\left({x}\right)\:{give} \\ $$$${I}\:=\int\:\mathrm{2}{ch}^{\mathrm{2}} \left({x}\right)\sqrt{\mathrm{2}}{sh}\left({x}\right)\sqrt{\mathrm{2}}{sh}\left({x}\right){dx} \\ $$$$=\mathrm{4}\:\int\:\:{ch}^{\mathrm{2}} {x}\:{sh}^{\mathrm{2}} {x}\:{dx}\:=\mathrm{4}\:\int\:\left({ch}\left({x}\right){sh}\left({x}\right)\right)^{\mathrm{2}} {dx} \\ $$$$=\mathrm{4}\:\int\left(\frac{\mathrm{1}}{\mathrm{2}}{sh}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} {dx}\:=\int\:\:{sh}^{\mathrm{2}} \left(\mathrm{2}{x}\right){dx}\:=\int\frac{{ch}\left(\mathrm{4}{x}\right)−\mathrm{1}}{\mathrm{2}}{dx} \\ $$$$=−\frac{{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}{sh}\left(\mathrm{4}{x}\right){dx}\:\:+{C} \\ $$$$=−\frac{{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{\frac{{e}^{\mathrm{4}{x}} −{e}^{−\mathrm{4}{x}} }{\mathrm{2}}\right\}\:+{C} \\ $$$$=−\frac{{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{16}}\left({e}^{\mathrm{4}{x}} −{e}^{−\mathrm{4}{x}} \right)\:+{C}\:\:{but}\:{x}={argch}\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$={ln}\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}+\sqrt{\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1}}\right)\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}+\sqrt{\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{16}}\left\{\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}+\sqrt{\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1}}\right)^{\mathrm{4}} −\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}+\sqrt{\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1}}\right)^{−\mathrm{4}} \right\}+{C} \\ $$$$= \\ $$
Commented by oyemi kemewari last updated on 02/Oct/19
thanks you slr
Commented by mathmax by abdo last updated on 03/Oct/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$

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