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Question Number 218674 by Spillover last updated on 14/Apr/25

Answered by Nicholas666 last updated on 14/Apr/25

$${\int }_0^1\left(\frac{2x^2-2x^4}{\sqrt{8+2x^2-x^4}}\right)\sqrt{\frac{2+x^2}{1-x^2}}dx$$$$solution;$$$${\int }_0^1\frac{2x^2(1-x^2)}{\sqrt{9-{(1-x^2)}^2}}\sqrt{\frac{2+x^2}{1-x^2}}dx={\int }_0^1\frac{2x^2\sqrt{1-x^2}\sqrt{2+x^2}}{\sqrt{9-{(1-x^2)}^2}}dx$$$$$$$$\mathrm{sub};1-x^2-y^2\mathrm{so}-2xdx=2ydy\mathrm{and}dx=\frac{-y}{1-y^2}dy;0⩽x⩽1;1⩾y⩾0$$$${\int }_1^0\frac{2(1-y^2)y^2}{\sqrt{9-y^4}}\sqrt{\frac{2+(1-y^2)}{y^2}}(-\frac{y}{\sqrt{1-y^2}})dy$$$$={\int }_0^1\frac{2(1-y^2)y^2}{\sqrt{9-y^4}}\frac{\sqrt{3-y^2}}{y}\frac{y}{\sqrt{1-y^2}}dy$$$$={\int }_0^1\frac{2y^2\sqrt{1-y^2}}{\sqrt{(3-y^2)(3+y^2)}}dy={\int }_0^1\frac{2y^2}{\sqrt{3+y^2}}dy$$$$$$$$\mathrm{sub}y=\sqrt{3}tan\varphi ,dy=\sqrt{3}{sec}^2\varphi d\varphi ;0⩽\varphi ⩽\pi /6$$$${\int }_0^{\pi /6}\frac{2(3{tan}^2\varphi }{\sqrt{3+3{tan}^2\varphi }}\sqrt{3}{sec}^2\varphi d\varphi$$$$={\int }_0^{\pi /6}\frac{6{tan}^2\varphi }{\sqrt{3}sec\varphi }\sqrt{3}{sec}^2\varphi d\varphi =6{\int }_0^{\pi /6}{tan}^2\varphi sec\varphi d\varphi$$$$=6{\int }_0^{\pi /6}({sec}^2\varphi -1)sec\varphi d\varphi =6{\int }_0^{\pi /6}({sec}^3\varphi -sec\varphi )d\varphi$$$$=6{[\frac{1}{2}(sec\varphi tan\varphi +ln\mid sec\varphi +tan\varphi \mid )-ln\mid sec\varphi +tan\varphi \mid ]}_0^{\pi /6}$$$$=3{[sec\varphi tan\varphi -ln\mid sec\varphi +tan\varphi \mid ]}_0^{\pi /6}$$$$=3[(\frac{2}{\sqrt{3}}\frac{1}{\sqrt{3}}-ln\mid \frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\mid )-(1.0-ln\mid 1+0\mid )]$$$$=3[\frac{2}{3}-ln\left(\frac{3}{\sqrt{3}}\right)-(0-0)]=3[\frac{2}{3}-ln\left(\sqrt{3}\right)]=2.3ln\left(\sqrt{3}\right)$$$$=2-\frac{3}{2}ln3✓$$$$$$

Answered by Spillover last updated on 14/Apr/25

Answered by Spillover last updated on 14/Apr/25

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