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Question Number 95951 by me2love2math last updated on 28/May/20

Commented by me2love2math last updated on 28/May/20

$$plshelpoutonthis$$

Commented by prakash jain last updated on 28/May/20

$$\int \frac{dx}{x^2\sqrt{4-x^2}}$$$$x=2\sin \theta$$$$dx=2\cos \theta d\theta$$$$\int \frac{2\cos \theta d\theta }{{\left(2\sin \theta \right)}^2\sqrt{4-{4\sin }^2\theta }}$$$$=\int \frac{2\cos \theta d\theta }{{4\sin }^2\theta \cdot 2\cos \theta }$$$$=\frac{1}{4}\int {\mathrm{cosec}}^2\theta d\theta$$$$=-\frac{1}{4}\cot \theta +C$$$$=-\frac{1}{4}\times \frac{\cos \theta }{\sin \theta }+C$$$$$$$$=-\frac{1}{4}\times \frac{\sqrt{1-{\sin }^2\theta }}{\sin \theta }+C$$$$=-\frac{1}{4}\times \frac{\sqrt{1-{\left(\frac{x}{2}\right)}^2}}{\frac{x}{2}}+C$$$$=-\frac{1}{4}\times \frac{\sqrt{4-x^2}}{x}+C$$$$$$$$Note:\cos \theta =\sqrt{1-{\sin }^2\theta }when-\frac{\pi }{2}<\theta <\frac{\pi }{2}$$

Commented by me2love2math last updated on 28/May/20

$$Thanks...plsQ2andQ3inparticular$$

Answered by abdomathmax last updated on 28/May/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{4-{\mathrm{x}}^2}}\mathrm{changement}\mathrm{x}=2\sin \theta \mathrm{give}$$$$\mathrm{I}=\int \frac{2\cos \theta \mathrm{d}\theta }{{4\sin }^2\theta \left(2\cos \theta \right)}=\int \frac{\mathrm{d}\theta }{4\times \frac{1-\cos \left(2\theta \right)}{2}}$$$$=\int \frac{\mathrm{d}\theta }{2-2\cos \left(2\theta \right)}{=}_{\tan \left(\theta \right)=\mathrm{t}}\int \frac{\mathrm{dt}}{(1+{\mathrm{t}}^2)(2-2\times \frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}$$$$=\int \frac{\mathrm{dt}}{2+{2\mathrm{t}}^2-2+{2\mathrm{t}}^2}=\int \frac{\mathrm{dt}}{{4\mathrm{t}}^2}=-\frac{1}{4\mathrm{t}}+\mathrm{c}$$$$=-\frac{1}{4\tan \theta }+\mathrm{c}=-\frac{\cos \theta }{4\sin \theta }+\mathrm{c}=-\frac{\sqrt{1-{\sin }^2\theta }}{2\mathrm{x}}+\mathrm{c}$$$$=-\frac{\sqrt{1-\frac{{\mathrm{x}}^2}{4}}}{2\mathrm{x}}+\mathrm{c}=-\frac{\sqrt{4-{\mathrm{x}}^2}}{4\mathrm{x}}+\mathrm{c}$$

Commented by me2love2math last updated on 28/May/20

$$thx...Q2andQ3iswhereihaveissues$$

Answered by Sourav mridha last updated on 28/May/20

$$\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{4-{\mathrm{x}}^2}}$$$$\mathrm{without}\mathrm{any}\mathrm{substitution}---$$$$\int \frac{\frac{1}{{\mathrm{x}}^3}}{\sqrt{\frac{4}{{\mathrm{x}}^2}-1}}\mathrm{dx}=\frac{1}{(-8)}\int \frac{\mathrm{d}(\frac{4}{{\mathrm{x}}^2}-1)}{\sqrt{\frac{4}{{\mathrm{x}}^2}-1}}$$$$=-\frac{1}{4}\sqrt{\frac{4}{{\mathrm{x}}^2}-1}+\mathrm{c}$$

Answered by john santu last updated on 29/May/20

$$\left(2\right)\underset{0}{\stackrel{1}{\int }}\left[\underset{\mathrm{x}}{\stackrel{\sqrt{\mathrm{x}}}{\int }}({60\mathrm{x}}^2\mathrm{y}-{40\mathrm{y}}^3)\mathrm{dy}\right]\mathrm{dx}$$$$\mathrm{the}\mathrm{inner}\mathrm{bracketed}\mathrm{integration}$$$$\mathrm{is}\mathrm{carried}\mathrm{out}\mathrm{first}\mathrm{and}\mathrm{evaluated}$$$$\mathrm{as}{[{30\mathrm{x}}^2{\mathrm{y}}^2-{10\mathrm{y}}^4]}_{\mathrm{x}}^{\sqrt{\mathrm{x}}}=$$$${30\mathrm{x}}^2(\mathrm{x}-{\mathrm{x}}^2)-10({\mathrm{x}}^2-{\mathrm{x}}^4)=$$$${30\mathrm{x}}^3-{30\mathrm{x}}^4-{10\mathrm{x}}^2+{10\mathrm{x}}^4=$$$${30\mathrm{x}}^3-{20\mathrm{x}}^4-{10\mathrm{x}}^2$$$$\mathrm{so}\mathrm{the}\mathrm{integration}\mathrm{becomes}$$$$\underset{0}{\stackrel{1}{\int }}({30\mathrm{x}}^3-{20\mathrm{x}}^4-{10\mathrm{x}}^2)\mathrm{dx}=$$$${[\frac{30}{4}{\mathrm{x}}^4-{4\mathrm{x}}^5-\frac{10}{3}{\mathrm{x}}^3]}_0^1=$$$$\frac{15}{2}-4-\frac{10}{3}=\frac{45-24-20}{6}$$$$=\frac{1}{6}.\mathrm{done}$$

Commented by me2love2math last updated on 29/May/20

$$thxsomuch$$

Answered by john santu last updated on 29/May/20

$$\left(3\right)\mathrm{f}(\mathrm{x},\mathrm{y})=2\mathrm{xy}-\frac{1}{2}({\mathrm{x}}^2+{\mathrm{y}}^2)+{\mathrm{x}}^2$$$$\mathrm{f}(\mathrm{x},\mathrm{y})=2\mathrm{xy}+\frac{1}{2}{\mathrm{x}}^2-\frac{1}{2}{\mathrm{y}}^2$$$$\mathrm{firstly}\mathrm{find}\mathrm{all}\mathrm{first}\mathrm{and}\mathrm{second}$$$$\mathrm{partial}\mathrm{derivatives}\mathrm{of}\mathrm{f}(\mathrm{x},\mathrm{y}).\mathrm{they}$$$$\mathrm{are}{f}_x=2y+x;f_y=2x-y$$$$f_{xx}=1;f_{yy}=-1$$$$f_{xy}=f_{yx}=2$$$$setting{f}_x=0,\mathrm{yields}x=-2y$$$$substitutioninto{f}_{y}yields-5y=0$$$$y=0.thissignalpossibleextrema$$$$at(0,0)\mathrm{as}\mathrm{x}=-2\mathrm{y}$$$$\mathrm{evaluate}\mathrm{D}\mathrm{at}(0,0)$$$$\mathrm{D}={4\mathrm{y}}^2-4\left(\frac{1}{2}\right)(-\frac{1}{2}{\mathrm{y}}^2)$$$$\mathrm{D}=0,\mathrm{there}\mathrm{is}\mathrm{an}\mathrm{extremum}$$$$\mathrm{and}\mathrm{as}{f}_{xx}>0itisaminimum$$$$thereforef(0,0)=0\mathrm{is}\mathrm{a}\mathrm{local}$$$$\mathrm{minimum}.\left(\mathrm{done}\right)$$

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