Question Number 218674 by Spillover last updated on 14/Apr/25
Answered by Nicholas666 last updated on 14/Apr/25
![∫_0 ^1 (((2x^2 −2x^4 )/( (√(8+2x^2 −x^4 )))))(√((2+x^2 )/(1−x^2 )))dx solution; ∫_0 ^1 ((2x^2 (1−x^2 ))/( (√(9−(1−x^2 )^2 )))) (√((2+x^2 )/(1−x^2 ))) dx= ∫_0 ^1 ((2x^2 (√(1−x^2 )) (√(2+x^2 )))/( (√(9−(1−x^2 )^2 )))) dx sub;1−x^2 −y^2 so −2xdx=2ydy and dx=((−y)/(1−y^2 ))dy; 0≤x≤1;1≥y≥0 ∫_1 ^0 ((2(1−y^2 )y^2 )/( (√(9−y^4 )))) (√((2+(1−y^2 ))/y^2 ))(−(y/( (√(1−y^2 )))))dy =∫_0 ^1 ((2(1−y^2 )y^2 )/( (√(9−y^4 )))) ((√(3−y^2 ))/y) (y/( (√(1−y^2 ))))dy =∫_0 ^1 ((2y^2 (√(1−y^2 )))/( (√((3−y^2 )(3+y^2 ))))) dy =∫_0 ^1 ((2y^2 )/( (√(3+y^2 ))))dy sub y=(√3) tanφ,dy=(√3) sec^2 φdφ ;0≤φ≤π/6 ∫_0 ^(π/6) ((2(3tan^2 φ)/( (√(3+3tan^2 φ))))(√3)sec^2 φdφ =∫_0 ^(π/6) ((6tan^2 φ)/( (√3)secφ))(√3)sec^2 φdφ=6∫_0 ^(π/6) tan^2 φsecφdφ =6∫_0 ^(π/6) (sec^2 φ−1)secφdφ=6∫_(0 ) ^(π/6) (sec^3 φ−secφ)dφ =6[(1/2)(secφtanφ+ln∣secφ+tanφ∣)−ln∣secφ+tanφ∣]_0 ^(π/6) =3[secφtanφ−ln∣secφ+tanφ∣]_0 ^(π/6) =3[((2/( (√3))) (1/( (√3))) − ln∣(2/( (√3)))+(1/( (√3)))∣)−(1.0−ln∣1+0∣)] =3[(2/3) − ln ((3/( (√3))))−(0−0)]=3[(2/3)−ln((√3))]=2.3ln((√3)) =2−(3/2)ln3 ✓](Q218696.png)
$$\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{4}} }{\:\sqrt{\mathrm{8}+\mathrm{2}{x}^{\mathrm{2}} −{x}^{\mathrm{4}} \:}}\right)\sqrt{\frac{\mathrm{2}+{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${solution}; \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{9}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }}\:\sqrt{\frac{\mathrm{2}+{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{9}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:}}\:{dx}\:\:\: \\ $$$$\: \\ $$$$\mathrm{sub};\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:\mathrm{so}\:−\mathrm{2}{xdx}=\mathrm{2}{ydy}\:\mathrm{and}\:{dx}=\frac{−{y}}{\mathrm{1}−{y}^{\mathrm{2}} }{dy};\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1};\mathrm{1}\geqslant{y}\geqslant\mathrm{0}\:\:\:\:\: \\ $$$$\int_{\mathrm{1}} ^{\mathrm{0}} \frac{\mathrm{2}\left(\mathrm{1}−{y}^{\mathrm{2}} \right){y}^{\mathrm{2}} }{\:\sqrt{\mathrm{9}−{y}^{\mathrm{4}} }}\:\sqrt{\frac{\mathrm{2}+\left(\mathrm{1}−{y}^{\mathrm{2}} \right)}{{y}^{\mathrm{2}} }}\left(−\frac{{y}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\right){dy}\:\: \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}\left(\mathrm{1}−{y}^{\mathrm{2}} \right){y}^{\mathrm{2}} }{\:\sqrt{\mathrm{9}−{y}^{\mathrm{4}} }}\:\frac{\sqrt{\mathrm{3}−{y}^{\mathrm{2}} }}{{y}}\:\frac{{y}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{dy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{y}^{\mathrm{2}} \sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{\:\sqrt{\left(\mathrm{3}−{y}^{\mathrm{2}} \right)\left(\mathrm{3}+{y}^{\mathrm{2}} \right)}}\:{dy}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{y}^{\mathrm{2}} }{\:\sqrt{\mathrm{3}+{y}^{\mathrm{2}} }}{dy}\:\:\: \\ $$$$ \\ $$$$\mathrm{sub}\:{y}=\sqrt{\mathrm{3}}\:{tan}\phi,{dy}=\sqrt{\mathrm{3}}\:{sec}^{\mathrm{2}} \phi{d}\phi\:\:;\mathrm{0}\leqslant\phi\leqslant\pi/\mathrm{6} \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \:\frac{\mathrm{2}\left(\mathrm{3}{tan}^{\mathrm{2}} \phi\right.}{\:\sqrt{\mathrm{3}+\mathrm{3}{tan}^{\mathrm{2}} \phi}}\sqrt{\mathrm{3}}{sec}^{\mathrm{2}} \phi{d}\phi \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \:\frac{\mathrm{6}{tan}^{\mathrm{2}} \phi}{\:\sqrt{\mathrm{3}}{sec}\phi}\sqrt{\mathrm{3}}{sec}^{\mathrm{2}} \phi{d}\phi=\mathrm{6}\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \:{tan}^{\mathrm{2}} \phi{sec}\phi{d}\phi\:\: \\ $$$$=\mathrm{6}\int_{\mathrm{0}} ^{\pi/\mathrm{6}} \left({sec}^{\mathrm{2}} \phi−\mathrm{1}\right){sec}\phi{d}\phi=\mathrm{6}\int_{\mathrm{0}\:} ^{\pi/\mathrm{6}} \left({sec}^{\mathrm{3}} \phi−{sec}\phi\right){d}\phi \\ $$$$=\mathrm{6}\left[\frac{\mathrm{1}}{\mathrm{2}}\left({sec}\phi{tan}\phi+{ln}\mid{sec}\phi+{tan}\phi\mid\right)−{ln}\mid{sec}\phi+{tan}\phi\mid\right]_{\mathrm{0}} ^{\pi/\mathrm{6}} \\ $$$$=\mathrm{3}\left[{sec}\phi{tan}\phi−{ln}\mid{sec}\phi+{tan}\phi\mid\right]_{\mathrm{0}} ^{\pi/\mathrm{6}} \\ $$$$=\mathrm{3}\left[\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:−\:{ln}\mid\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mid\right)−\left(\mathrm{1}.\mathrm{0}−{ln}\mid\mathrm{1}+\mathrm{0}\mid\right)\right] \\ $$$$=\mathrm{3}\left[\frac{\mathrm{2}}{\mathrm{3}}\:−\:{ln}\:\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}}\right)−\left(\mathrm{0}−\mathrm{0}\right)\right]=\mathrm{3}\left[\frac{\mathrm{2}}{\mathrm{3}}−{ln}\left(\sqrt{\mathrm{3}}\right)\right]=\mathrm{2}.\mathrm{3}{ln}\left(\sqrt{\mathrm{3}}\right) \\ $$$$=\mathrm{2}−\frac{\mathrm{3}}{\mathrm{2}}{ln}\mathrm{3}\:\checkmark \\ $$$$ \\ $$
Answered by Spillover last updated on 14/Apr/25
Answered by Spillover last updated on 14/Apr/25
