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Question Number 54919 by maxmathsup by imad last updated on 14/Feb/19
calculate ∫_1 ^(√2) (x^3 +1)(√(x^2 −1))dx
$${calculate}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \left({x}^{\mathrm{3}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$
Commented by Abdo msup. last updated on 15/Feb/19
let I =∫_1 ^(√2) (x^3 +1)(√(x^2 −1)) dx changement x=cht gove I =∫_(argch(1)) ^(argch((√2))) (ch^3 t+1)sh^2 (t) dt =∫_0 ^(ln(1+(√2))) ch^3 t sh^2 t dt +∫_0 ^(ln(1+(√2))) sh^2 t dt ∫_0 ^(ln(1+(√2))) ((ch(2t)−1)/2)dt =(1/4)[sh(2t)]_0 ^(ln(1+(√2))) −((ln(1+(√2)))/2) =(1/8)[ e^(2t) −e^(−2t) ]_0 ^(ln(1+(√2))) −((ln(1+(√2)))/2) =(1/8){ (1+(√2))^2 −(1/((1+(√2))^2 ))}−((ln(1+(√2)))/2) also we have ch^3 tsh^2 t =(chtsht)^2 cht =(1/4)sh^2 (2t)ch(t) =(1/4)(((ch(4t)−1)/2))ch(t) =(1/8)(((e^(4t) +e^(−4t) )/2) −1)(((e^t +e^(−t) )/2)) =(1/(32))(e^(4t) +e^(−4t) −2)(e^t +e^(−t) ) =(1/(32))(e^(5t ) +e^(3t ) +e^(−3t) +e^(−5t) −2e^t −2e^(−t) ) ⇒ ∫_0 ^(ln(1+(√2))) ch^3 t sh^2 t dt =(1/(32)) ∫_0 ^(ln(1+(√2))) ( e^(5t) +e^(−5t) +e^(3t) +e^(−3t) −2e^t −2e^(−t) )dt =(1/(32))[(1/5)e^(5t) −(1/5)e^(−5t) +(1/3)e^(3t) −(1/3)e^(−3t) −2e^t +2e^(−t) ]_0 ^(ln(1+(√2))) =(1/(32)){ (1/5)( (1+(√2))^2 −(1/((1+(√2))^2 )))+(1/3)((1+(√2))^3 −(1/((1+(√2))^3 ))) −2( 1+(√2)−(1/(1+(√2))))} so the value of I is determined
$${let}\:{I}\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \left({x}^{\mathrm{3}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:{dx}\:{changement}\:{x}={cht}\:{gove} \\ $$$${I}\:=\int_{{argch}\left(\mathrm{1}\right)} ^{{argch}\left(\sqrt{\mathrm{2}}\right)} \left({ch}^{\mathrm{3}} {t}+\mathrm{1}\right){sh}^{\mathrm{2}} \left({t}\right)\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} {ch}^{\mathrm{3}} {t}\:{sh}^{\mathrm{2}} {t}\:{dt}\:+\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} {sh}^{\mathrm{2}} {t}\:{dt} \\ $$$$\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}{dt}\:=\frac{\mathrm{1}}{\mathrm{4}}\left[{sh}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$$−\frac{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{8}}\left[\:{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} \right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:−\frac{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left\{\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right\}−\frac{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}} \\ $$$${also}\:{we}\:{have}\:{ch}^{\mathrm{3}} {tsh}^{\mathrm{2}} {t}\:=\left({chtsht}\right)^{\mathrm{2}} {cht} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{sh}^{\mathrm{2}} \left(\mathrm{2}{t}\right){ch}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{ch}\left(\mathrm{4}{t}\right)−\mathrm{1}}{\mathrm{2}}\right){ch}\left({t}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{{e}^{\mathrm{4}{t}} \:+{e}^{−\mathrm{4}{t}} }{\mathrm{2}}\:−\mathrm{1}\right)\left(\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\left({e}^{\mathrm{4}{t}} \:+{e}^{−\mathrm{4}{t}} −\mathrm{2}\right)\left({e}^{{t}} \:+{e}^{−{t}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\left({e}^{\mathrm{5}{t}\:} \:+{e}^{\mathrm{3}{t}\:} \:+{e}^{−\mathrm{3}{t}} \:+{e}^{−\mathrm{5}{t}} \:−\mathrm{2}{e}^{{t}} −\mathrm{2}{e}^{−{t}} \right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} {ch}^{\mathrm{3}} {t}\:{sh}^{\mathrm{2}} {t}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \left(\:{e}^{\mathrm{5}{t}} \:+{e}^{−\mathrm{5}{t}} \:+{e}^{\mathrm{3}{t}} \:+{e}^{−\mathrm{3}{t}} −\mathrm{2}{e}^{{t}} −\mathrm{2}{e}^{−{t}} \right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\left[\frac{\mathrm{1}}{\mathrm{5}}{e}^{\mathrm{5}{t}} −\frac{\mathrm{1}}{\mathrm{5}}{e}^{−\mathrm{5}{t}} \:+\frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{3}{t}} −\frac{\mathrm{1}}{\mathrm{3}}{e}^{−\mathrm{3}{t}} −\mathrm{2}{e}^{{t}} \:+\mathrm{2}{e}^{−{t}} \right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\left\{\:\frac{\mathrm{1}}{\mathrm{5}}\left(\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{\mathrm{3}}\left(\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{3}} }\right)\right. \\ $$$$\left.−\mathrm{2}\left(\:\mathrm{1}+\sqrt{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}}}\right)\right\}\:\:{so}\:{the}\:{value}\:{of}\:{I}\:{is}\:{determined} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19
∫x×x^2 (√(x^2 −1)) dx+∫(√(x^2 −1)) dx t^2 =x^2 −1→tdt=xdx ∫(t^2 +1)t×tdt+∫(√(x^2 −1)) dx ∫t^4 +t^2 dt+∫(√(x^2 −1)) dx (t^5 /5)+(t^3 /3)+(x/2)(√(x^2 −1)) −(1/2)ln(x+(√(x^2 −1)) )+c (((x^2 −1)^(5/2) )/5)+(((x^2 −1)^(3/2) )/3)+(x/2)(√(x^2 −1)) −(1/2)ln(x+(√(x^2 −1)) )+c so answer is ∣(((x^2 −1)^(5/2) )/5)+(((x^2 −1)^(3/2) )/3)+(x/2)(√(x^2 −1)) −(1/2)ln(x+(√(x^2 −1)) )∣_1 ^(√2) [(1/5)+(1/3)+((√2)/2)(√(2−1)) −(1/2)ln((√2) +1)] (8/(15))+((√2)/2)−(1/2)ln((√2) +1 )
$$\int{x}×{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:{dx}+\int\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:{dx} \\ $$$${t}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{1}\rightarrow{tdt}={xdx} \\ $$$$\int\left({t}^{\mathrm{2}} +\mathrm{1}\right){t}×{tdt}+\int\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:{dx} \\ $$$$\int{t}^{\mathrm{4}} +{t}^{\mathrm{2}} \:{dt}+\int\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:{dx} \\ $$$$\frac{{t}^{\mathrm{5}} }{\mathrm{5}}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\right)+{c} \\ $$$$\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{5}}+\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}+\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\right)+{c} \\ $$$${so}\:{answer}\:{is} \\ $$$$\mid\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{5}}+\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}+\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\right)\mid_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\mathrm{2}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{2}}\:+\mathrm{1}\right)\right] \\ $$$$\frac{\mathrm{8}}{\mathrm{15}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{2}}\:+\mathrm{1}\:\right) \\ $$$$ \\ $$

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