Question Number 213369 by Frix last updated on 03/Nov/24
$$\mathrm{Old}\:\mathrm{question}\:\mathrm{203835} \\ $$$$\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\int}}\frac{\sqrt{\mathrm{6}−\sqrt{\mathrm{25}{x}^{\mathrm{4}} −\mathrm{50}{x}^{\mathrm{2}} +\mathrm{36}}}}{\:\sqrt{\mathrm{5}}}{dx}=? \\ $$
Answered by Frix last updated on 03/Nov/24
![∫_0 ^(√2) ((√(6−(√(25x^4 −50x^2 +36))))/( (√5)))dx =^([t=(5/( (√(11))))(x^2 −1)]) =((√(11))/(10))∫_(−(5/( (√(11))))) ^(5/( (√(11)))) ((√(6−(√(11(t^2 +1)))))/( (√((√(11))t+5))))dt =^([u=t+(√(t^2 +1))]) =((√(11))/(20))∫_(1/( (√(11)))) ^(√(11)) (((u^2 +1)(√(−(√(11))u^2 +12u−(√(11)))))/(u^2 (√((√(11))u^2 +10u−(√(11))))))du= =((√(11))/(20))∫_(1/( (√(11)))) ^(√(11)) (((u^2 +1)(√(((√(11))−u)(u−(1/( (√(11))))))))/(u^2 (√(((√(11))+u)(u−(1/( (√(11)))))))))du= =((√(11))/(20))∫_(1/( (√(11)))) ^(√(11)) (((u^2 +1)(√((√(11))−u)))/(u^2 (√((√(11))+u))))du =^([v=((√((√(11))−u))/( (√((√(11))+u))))]) =(4/5)∫_( 0) ^((√5)/( (√6))) ((v^2 (3v^4 −5v^2 +3))/((v−1)^2 (v+1)^2 (v^2 +1)^2 ))dv= =(1/(20))∫_0 ^((√5)/( (√6))) ((1/((v−1)^2 ))+(1/((v+1)^2 ))+(1/(v−1))−(1/(v+1)))dv+ +((11)/5)∫_0 ^((√5)/( (√6))) ((1/(v^2 +1))−(1/((v^2 +1)^2 )))dv= ... =[−((v(6v^2 −5))/(5(v^4 −1)))+(1/(20))ln ∣((v−1)/(v+1))∣ +((11)/(10))tan^(−1) v]_0 ^((√5)/( (√6))) = =(1/(10))ln ((√6)−(√5)) +((11)/(10))tan^(−1) ((√5)/( (√6)))](https://www.tinkutara.com/question/Q213370.png)
$$\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\int}}\frac{\sqrt{\mathrm{6}−\sqrt{\mathrm{25}{x}^{\mathrm{4}} −\mathrm{50}{x}^{\mathrm{2}} +\mathrm{36}}}}{\:\sqrt{\mathrm{5}}}{dx}\:\overset{\left[{t}=\frac{\mathrm{5}}{\:\sqrt{\mathrm{11}}}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\right]} {=} \\ $$$$=\frac{\sqrt{\mathrm{11}}}{\mathrm{10}}\underset{−\frac{\mathrm{5}}{\:\sqrt{\mathrm{11}}}} {\overset{\frac{\mathrm{5}}{\:\sqrt{\mathrm{11}}}} {\int}}\frac{\sqrt{\mathrm{6}−\sqrt{\mathrm{11}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}}}{\:\sqrt{\sqrt{\mathrm{11}}{t}+\mathrm{5}}}{dt}\:\overset{\left[{u}={t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right]} {=} \\ $$$$=\frac{\sqrt{\mathrm{11}}}{\mathrm{20}}\underset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}} {\overset{\sqrt{\mathrm{11}}} {\int}}\frac{\left({u}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{−\sqrt{\mathrm{11}}{u}^{\mathrm{2}} +\mathrm{12}{u}−\sqrt{\mathrm{11}}}}{{u}^{\mathrm{2}} \sqrt{\sqrt{\mathrm{11}}{u}^{\mathrm{2}} +\mathrm{10}{u}−\sqrt{\mathrm{11}}}}{du}= \\ $$$$=\frac{\sqrt{\mathrm{11}}}{\mathrm{20}}\underset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}} {\overset{\sqrt{\mathrm{11}}} {\int}}\frac{\left({u}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\left(\sqrt{\mathrm{11}}−{u}\right)\left({u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}\right)}}{{u}^{\mathrm{2}} \sqrt{\left(\sqrt{\mathrm{11}}+{u}\right)\left({u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}\right)}}{du}= \\ $$$$=\frac{\sqrt{\mathrm{11}}}{\mathrm{20}}\underset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}} {\overset{\sqrt{\mathrm{11}}} {\int}}\frac{\left({u}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\sqrt{\mathrm{11}}−{u}}}{{u}^{\mathrm{2}} \sqrt{\sqrt{\mathrm{11}}+{u}}}{du}\:\overset{\left[{v}=\frac{\sqrt{\sqrt{\mathrm{11}}−{u}}}{\:\sqrt{\sqrt{\mathrm{11}}+{u}}}\right]} {=} \\ $$$$=\frac{\mathrm{4}}{\mathrm{5}}\underset{\:\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{6}}}} {\int}}\frac{{v}^{\mathrm{2}} \left(\mathrm{3}{v}^{\mathrm{4}} −\mathrm{5}{v}^{\mathrm{2}} +\mathrm{3}\right)}{\left({v}−\mathrm{1}\right)^{\mathrm{2}} \left({v}+\mathrm{1}\right)^{\mathrm{2}} \left({v}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dv}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{20}}\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{6}}}} {\int}}\left(\frac{\mathrm{1}}{\left({v}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({v}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{{v}−\mathrm{1}}−\frac{\mathrm{1}}{{v}+\mathrm{1}}\right){dv}+ \\ $$$$\:\:\:\:\:+\frac{\mathrm{11}}{\mathrm{5}}\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{6}}}} {\int}}\left(\frac{\mathrm{1}}{{v}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\left({v}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\right){dv}= \\ $$$$… \\ $$$$=\left[−\frac{{v}\left(\mathrm{6}{v}^{\mathrm{2}} −\mathrm{5}\right)}{\mathrm{5}\left({v}^{\mathrm{4}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{20}}\mathrm{ln}\:\mid\frac{{v}−\mathrm{1}}{{v}+\mathrm{1}}\mid\:+\frac{\mathrm{11}}{\mathrm{10}}\mathrm{tan}^{−\mathrm{1}} \:{v}\right]_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{6}}}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\mathrm{ln}\:\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{5}}\right)\:+\frac{\mathrm{11}}{\mathrm{10}}\mathrm{tan}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{6}}} \\ $$
Answered by MrGaster last updated on 03/Nov/24
$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{\sqrt{\mathrm{6}−\sqrt{\left(\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}\right)^{\mathrm{2}} }}}{\:\sqrt{\mathrm{5}}}{dx}=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{\sqrt{\mathrm{6}−\mid\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}\mid}}{\:\sqrt{\mathrm{5}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{\sqrt{\mathrm{6}−\left(\mathrm{6}−\mathrm{5}{x}^{\mathrm{2}} \right)}}{\:\sqrt{\mathrm{5}}}{dx}\:\:\:\:\:\:\mathrm{for}\:\mathrm{0}\leq{x}\leq\sqrt{\mathrm{2}},\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}\leq\mathrm{0} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} {x}\:{dx}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mid_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \\ $$$$=\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{0}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}} \\ $$$$=\mathrm{1}\:\mathrm{or}\:\approx\mathrm{0}.\mathrm{65942035657438469264675199091487656785970133168787978938786781873469280290781916555575665168451357245368318166250688730704685276141919488865053159337}… \\ $$
Commented by Frix last updated on 04/Nov/24
$$\mathrm{Again}:\:\mathrm{this}\:\mathrm{is}\:\mathrm{wrong} \\ $$$$\mathrm{25}{x}^{\mathrm{4}} −\mathrm{50}{x}^{\mathrm{2}} +\mathrm{36}\neq\left(\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}\right)^{\mathrm{2}} =\mathrm{25}{x}^{\mathrm{4}} −\mathrm{60}{x}^{\mathrm{2}} +\mathrm{36} \\ $$$$\mathrm{And}\:\frac{\mathrm{2}}{\mathrm{1}}\neq\mathrm{1}\neq.\mathrm{0659}… \\ $$$$\mathrm{I}\:\mathrm{doubt}\:\mathrm{you}\:\mathrm{have}\:\mathrm{any}\:\mathrm{clue}\:\mathrm{at}\:\mathrm{all} \\ $$