Question Number 203836 by patrice last updated on 29/Jan/24
Answered by Frix last updated on 30/Jan/24
![In 3 steps: 1. t=x^2 −1 2. u=((5t+(√(25t^2 +11)))/( 11)) 3. v=((√(1+u))/( (√(1−u)))) ⇒ (4/5) ∫_((√6)/( (√5))) ^∞ ((v^2 (3v^4 −5v^2 +3))/((v^4 −1)^2 ))dv= =[((v(5v^2 −6))/(5(v^4 −1)))+((ln ((v+1)/(v−1)))/(20))+((11tan^(−1) v)/(10))]_((√6)/( (√5))) ^∞ = =((11π)/(20))−((ln ((√6)+(√5)))/(10))−((11tan^(−1) ((√6)/( (√5))))/(10))](https://www.tinkutara.com/question/Q203840.png)
$$\mathrm{In}\:\mathrm{3}\:\mathrm{steps}: \\ $$$$\mathrm{1}.\:{t}={x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{2}.\:{u}=\frac{\mathrm{5}{t}+\sqrt{\mathrm{25}{t}^{\mathrm{2}} +\mathrm{11}}}{\:\mathrm{11}} \\ $$$$\mathrm{3}.\:{v}=\frac{\sqrt{\mathrm{1}+{u}}}{\:\sqrt{\mathrm{1}−{u}}}\:\Rightarrow \\ $$$$\frac{\mathrm{4}}{\mathrm{5}}\:\underset{\frac{\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{5}}}} {\overset{\infty} {\int}}\:\frac{{v}^{\mathrm{2}} \left(\mathrm{3}{v}^{\mathrm{4}} −\mathrm{5}{v}^{\mathrm{2}} +\mathrm{3}\right)}{\left({v}^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{2}} }{dv}= \\ $$$$=\left[\frac{{v}\left(\mathrm{5}{v}^{\mathrm{2}} −\mathrm{6}\right)}{\mathrm{5}\left({v}^{\mathrm{4}} −\mathrm{1}\right)}+\frac{\mathrm{ln}\:\frac{{v}+\mathrm{1}}{{v}−\mathrm{1}}}{\mathrm{20}}+\frac{\mathrm{11tan}^{−\mathrm{1}} \:{v}}{\mathrm{10}}\right]_{\frac{\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{5}}}} ^{\infty} = \\ $$$$=\frac{\mathrm{11}\pi}{\mathrm{20}}−\frac{\mathrm{ln}\:\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{5}}\right)}{\mathrm{10}}−\frac{\mathrm{11tan}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{5}}}}{\mathrm{10}} \\ $$
Commented by patrice last updated on 30/Jan/24
je ne sais pas bien merci
Answered by Sutrisno last updated on 30/Jan/24
$$ \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{\sqrt{\mathrm{6}−\left(\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}\right)^{\mathrm{2}} }}{\:\sqrt{\mathrm{5}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{\sqrt{\mathrm{12}−\mathrm{5}{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{5}}}{dx} \\ $$$${misal}\:\sqrt{\mathrm{5}}{x}=\sqrt{\mathrm{12}}{sin}\theta\rightarrow\sqrt{\mathrm{5}}{dx}=\sqrt{\mathrm{12}}{cos}\theta{d}\theta \\ $$$$=\int\frac{\sqrt{\mathrm{12}−\mathrm{12}{sin}^{\mathrm{2}} \theta}}{\:\sqrt{\mathrm{5}}}.\frac{\sqrt{\mathrm{12}}{cos}\theta{d}\theta}{\:\sqrt{\mathrm{5}}} \\ $$$$=\frac{\mathrm{12}}{\mathrm{5}}\int{cos}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\frac{\mathrm{12}}{\mathrm{5}}\int\frac{{cos}\mathrm{2}\theta+\mathrm{1}}{\mathrm{2}}{d}\theta \\ $$$$=\frac{\mathrm{6}}{\mathrm{5}}\left(\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}\theta+\theta\right) \\ $$$$=\frac{\mathrm{6}}{\mathrm{5}}\left({sin}\theta{cos}\theta+\theta\right) \\ $$$$=\frac{\mathrm{6}}{\mathrm{5}}\left(\frac{\sqrt{\mathrm{5}}{x}}{\:\sqrt{\mathrm{12}}}.\frac{\sqrt{\mathrm{12}−\mathrm{5}{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{12}}}+{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{5}}{x}}{\:\sqrt{\mathrm{12}}}\right)\right)\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\mid}} \\ $$$$=\frac{\mathrm{6}}{\mathrm{5}}\left(\frac{\sqrt{\mathrm{10}}}{\:\sqrt{\mathrm{12}}}.\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{12}}}+{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{10}}}{\:\sqrt{\mathrm{12}}}\right)\right) \\ $$$$=\frac{\mathrm{6}}{\mathrm{5}}\left(\frac{\sqrt{\mathrm{5}}}{\:\mathrm{6}}+{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{30}}}{\:\mathrm{6}}\right)\right) \\ $$$$ \\ $$
Commented by Frix last updated on 30/Jan/24
$$\mathrm{Wrong}\:\mathrm{because} \\ $$$$\sqrt{\mathrm{25}{x}^{\mathrm{4}} −\mathrm{50}{x}^{\mathrm{2}} +\mathrm{36}}\neq\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6} \\ $$$$\mathrm{And}\:\mathrm{we}\:\mathrm{see}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{a}\:\mathrm{typo}\:\mathrm{because}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{upper}\:\mathrm{border}\:\sqrt{\mathrm{2}}\:\mathrm{where}\:\mathrm{the}\:\mathrm{given}\:\mathrm{function} \\ $$$$\mathrm{equals}\:\mathrm{zero}. \\ $$
Commented by Frix last updated on 30/Jan/24
![∫_0 ^b ((√(6−(√(25x^4 −60x^2 +36))))/( (√5)))dx= =∫_0 ^b ((√(6−∣5x^2 −6∣))/( (√5)))dx= =∫_0 ^((√(30))/5) xdx+∫_((√(30))/( 5)) ^b ((√(12−5x^2 ))/( (√5)))dx= =[(x^2 /2)]_0 ^((√(30))/5) +[((x(√(12−5x^2 )))/(2(√5)))+((6sin^(−1) (((√(15))x)/6))/5)]_((√(30))/5) ^b This gives: b=(√2) ((√5)/5)−((3π)/(10))+((6sin^(−1) ((√(30))/6))/5) b=((2(√(15)))/5) [zero of the function] ((3π)/(10))](https://www.tinkutara.com/question/Q203855.png)
$$\underset{\mathrm{0}} {\overset{{b}} {\int}}\frac{\sqrt{\mathrm{6}−\sqrt{\mathrm{25}{x}^{\mathrm{4}} −\mathrm{60}{x}^{\mathrm{2}} +\mathrm{36}}}}{\:\sqrt{\mathrm{5}}}{dx}= \\ $$$$=\underset{\mathrm{0}} {\overset{{b}} {\int}}\frac{\sqrt{\mathrm{6}−\mid\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}\mid}}{\:\sqrt{\mathrm{5}}}{dx}= \\ $$$$=\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{30}}}{\mathrm{5}}} {\int}}{xdx}+\underset{\frac{\sqrt{\mathrm{30}}}{\:\mathrm{5}}} {\overset{{b}} {\int}}\frac{\sqrt{\mathrm{12}−\mathrm{5}{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{5}}}{dx}= \\ $$$$=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{30}}}{\mathrm{5}}} +\left[\frac{{x}\sqrt{\mathrm{12}−\mathrm{5}{x}^{\mathrm{2}} }}{\mathrm{2}\sqrt{\mathrm{5}}}+\frac{\mathrm{6sin}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{15}}{x}}{\mathrm{6}}}{\mathrm{5}}\right]_{\frac{\sqrt{\mathrm{30}}}{\mathrm{5}}} ^{{b}} \\ $$$$\mathrm{This}\:\mathrm{gives}: \\ $$$${b}=\sqrt{\mathrm{2}} \\ $$$$\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}−\frac{\mathrm{3}\pi}{\mathrm{10}}+\frac{\mathrm{6sin}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{30}}}{\mathrm{6}}}{\mathrm{5}} \\ $$$${b}=\frac{\mathrm{2}\sqrt{\mathrm{15}}}{\mathrm{5}}\:\left[\mathrm{zero}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\right] \\ $$$$\frac{\mathrm{3}\pi}{\mathrm{10}} \\ $$