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Question Number 94383 by niroj last updated on 18/May/20

$$\mathbf{Integrate}:$$$$\left(\mathbf{i}\right).\int \frac{1}{1+{\mathbf{x}}^4}\mathbf{dx}$$$$\left(\mathbf{ii}\right).{\int }_{\beta }^{\alpha }\sqrt{(\mathbf{x}-\mathit{\alpha })(\beta -\mathbf{x})}\mathbf{dx}$$$$\left(\mathbf{iii}\right).{\int }_0^1{\int }_0^{{\mathbf{x}}^2}{\mathbf{e}}^{\mathbf{y}/\mathbf{x}}\mathbf{dx}\mathbf{dy}$$

Answered by Mr.D.N. last updated on 18/May/20

$$\left(\mathrm{i}\right).\int \frac{1}{1+{\mathrm{x}}^4}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{{\mathrm{x}}^2+1-{\mathrm{x}}^2+1}{1+{\mathrm{x}}^4}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{{\mathrm{x}}^2+1}{1+{\mathrm{x}}^4}\mathrm{dx}-\frac{1}{2}\int \frac{{\mathrm{x}}^2-1}{1+{\mathrm{x}}^4}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{{\mathrm{x}}^2(1+\frac{1}{{\mathrm{x}}^2})}{{\mathrm{x}}^2({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2})}\mathrm{dx}-\frac{1}{2}\int \frac{{\mathrm{x}}^2(1-\frac{1}{{\mathrm{x}}^2})}{{\mathrm{x}}^2({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2})}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{(1+\frac{1}{{\mathrm{x}}^2})\mathrm{dx}}{{(\mathrm{x}-\frac{1}{\mathrm{x}})}^2+2}-\frac{1}{2}\int \frac{(1-\frac{1}{{\mathrm{x}}^2})\mathrm{dx}}{{(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-2}$$$$\mathrm{Put},(\mathrm{x}-\frac{1}{\mathrm{x}})=\mathrm{t}\mathrm{\&}(\mathrm{x}+\frac{1}{\mathrm{x}})=\mathrm{m}$$$$(1+\frac{1}{{\mathrm{x}}^2})\mathrm{dx}=\mathrm{dt},(1-\frac{1}{{\mathrm{x}}^2})\mathrm{dx}=\mathrm{dm}$$$$=\frac{1}{2}\int \frac{1}{{\left(\mathrm{t}\right)}^2+{\left(\sqrt{2}\right)}^2}\mathrm{dt}-\frac{1}{2}\int \frac{1}{{\left(\mathrm{m}\right)}^2-{\left(\sqrt{2}\right)}^2}\mathrm{dm}$$$$=\frac{1}{2}.\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{\mathrm{t}}{\sqrt{2}}-\frac{1}{2}.\frac{1}{2\sqrt{2}}\log \frac{\mathrm{m}-\sqrt{2}}{\mathrm{m}+\sqrt{2}}+\mathrm{C}$$$$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\frac{\mathrm{x}-\frac{1}{\mathrm{x}}}{\sqrt{2}}-\frac{1}{4\sqrt{2}}\log \frac{\mathrm{x}+\frac{1}{\mathrm{x}}-\sqrt{2}}{\mathrm{x}+\frac{1}{\mathrm{x}}+\sqrt{2}}+\mathrm{C}$$$$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\frac{{\mathrm{x}}^2-1}{\sqrt{2}\mathrm{x}}-\frac{1}{4\sqrt{2}}\log \frac{{\mathrm{x}}^2-\sqrt{2}\mathrm{x}+1}{{\mathrm{x}}^2+\sqrt{2}\mathrm{x}+1}+\mathrm{C}//.$$$$$$$$$$

Commented by niroj last updated on 18/May/20

Excellent ����������

Commented by niroj last updated on 18/May/20

������

Commented by peter frank last updated on 18/May/20

$$thankyou$$

Answered by Ar Brandon last updated on 18/May/20

$$\mathrm{iii}\mathrm{\setminus }{\int }_0^1{\int }_0^{{\mathrm{x}}^2}{\mathrm{e}}^{\mathrm{y}/\mathrm{x}}\mathrm{dxdy}={\int }_0^1\mathrm{x}{\left[{\mathrm{e}}^{\mathrm{y}/\mathrm{x}}\right]}_0^{{\mathrm{x}}^2}\mathrm{dx}={\int }_0^1\mathrm{x}({\mathrm{e}}^{\mathrm{x}}-1)\mathrm{dx}={\int }_0^1\mathrm{x}({\mathrm{e}}^{\mathrm{x}}-1)\mathrm{dx}$$$${[\mathrm{x}\int ({\mathrm{e}}^{\mathrm{x}}-1)\mathrm{dx}]}_0^1-{\int }_0^1\{\frac{\mathrm{dx}}{\mathrm{dx}}\int ({\mathrm{e}}^{\mathrm{x}}-1)\mathrm{dx}\}\mathrm{dx}={\left[\mathrm{x}({\mathrm{e}}^{\mathrm{x}}-\mathrm{x})\right]}_0^1-{\int }_0^1({\mathrm{e}}^{\mathrm{x}}-\mathrm{x})\mathrm{dx}$$$$=\mathrm{e}-1-{[{\mathrm{e}}^{\mathrm{x}}-\frac{{\mathrm{x}}^2}{2}]}_0^1=\mathrm{e}-1-\mathrm{e}+\frac{1}{2}+1=\frac{1}{2}$$

Commented by niroj last updated on 18/May/20

excellent dear..��������

Commented by Ar Brandon last updated on 18/May/20

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Answered by mathmax by abdo last updated on 18/May/20

$$\mathrm{I}={\int }_{\beta }^{\alpha }\sqrt{(\mathrm{x}-\alpha )(\beta -\mathrm{x})}\mathrm{dx}\Rightarrow \mathrm{I}={\int }_{\beta }^{\alpha }\sqrt{(\alpha -\mathrm{x})(\mathrm{x}-\beta )}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}$$$$\mathrm{changement}\sqrt{\alpha -\mathrm{x}}=\mathrm{t}\Rightarrow \alpha -\mathrm{x}={\mathrm{t}}^2\Rightarrow \mathrm{x}=\alpha -{\mathrm{t}}^2\Rightarrow$$$$\mathrm{I}={\int }_{\sqrt{\alpha -\beta }}^0\mathrm{t}\sqrt{\alpha -{\mathrm{t}}^2-\beta }(-2\mathrm{t})\mathrm{dt}=2{\int }_0^{\sqrt{\alpha -\beta }}{\mathrm{t}}^2\sqrt{\alpha -\beta -{\mathrm{t}}^2}\mathrm{dt}$$$${=}_{\mathrm{t}=\sqrt{\alpha -\beta }\mathrm{sinu}}2{\int }_0^{\sqrt{\alpha -\beta }}(\alpha -\beta ){\sin }^2\mathrm{u}\sqrt{\alpha -\beta }\mathrm{cosu}\sqrt{\alpha -\beta }\mathrm{cosu}\mathrm{du}$$$$=2{(\alpha -\beta )}^2{\int }_0^{\frac{\pi }{2}}{\sin }^2\mathrm{u}{\cos }^2\mathrm{u}\mathrm{du}=2{(\alpha -\beta )}^2\times \frac{1}{4}{\int }_0^{\frac{\pi }{2}}{\sin }^2\left(2\mathrm{u}\right)\mathrm{du}$$$$=\frac{{(\alpha -\beta )}^2}{4}{\int }_0^{\frac{\pi }{2}}(1-\cos \left(4\mathrm{u}\right))\mathrm{du}$$$$=\frac{\pi }{8}{(\alpha -\beta )}^2-\frac{{(\alpha -\beta )}^2}{2}{\left[\frac{1}{4}\sin \left(4\mathrm{u}\right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{8}{(\alpha -\beta )}^2$$

Commented by niroj last updated on 18/May/20

it's to be great����������

Answered by Mr.D.N. last updated on 18/May/20

$$\left(\mathrm{ii}\right).\mathrm{I}={\int }_{\beta }^{\alpha }\sqrt{(\mathrm{x}-\alpha )(\beta -\mathrm{x})}\mathrm{dx}$$$$={\int }_{\beta }^{\alpha }\sqrt{-{\mathrm{x}}^2+(\alpha +\beta )\mathrm{x}-\alpha \beta }\mathrm{dx}$$$$=={\int }_{\beta }^{\alpha }\sqrt{{\left(\frac{\beta -\alpha }{2}\right)}^2-\alpha \beta -[{\mathrm{x}}^2-(\alpha +\beta )\mathrm{x}+{\left(\frac{\beta +\alpha }{2}\right)}^2]}\mathrm{dx}$$$$={\int }_{\beta }^{\alpha }\sqrt{{\left(\frac{\beta -\alpha }{2}\right)}^2-{(\mathrm{x}-\frac{\beta +\alpha }{2})}^2}\mathrm{dx}$$$$\mathrm{Put},\frac{\beta -\alpha }{2}=\mathrm{a},\mathrm{and}\mathrm{x}-\frac{\beta +\alpha }{2}=\mathrm{y}$$$$\mathrm{Then}\mathrm{dx}=\mathrm{dy}$$$$\mathrm{When}\mathrm{x}=\alpha ,\mathrm{y}=\alpha -\frac{\beta +\alpha }{2}=\frac{\alpha -\beta }{2}=-\mathrm{a}$$$$\mathrm{When}\mathrm{x}=\beta ,\mathrm{y}=\beta -\frac{\beta +\alpha }{2}=\frac{\beta -\alpha }{2}=\mathrm{a}$$$$\therefore \mathrm{I}={\int }_{-\mathrm{a}}^{\mathrm{a}}\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}\mathrm{dy}$$$$={[\frac{\mathrm{y}\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{2}+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\frac{\mathrm{y}}{\mathrm{a}}]}_{-\mathrm{a}}^{\mathrm{a}}$$$$=\{0+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\left(1\right)\}-\{0+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}(-1)\}$$$$=\frac{{\mathrm{a}}^2}{2}.\frac{\pi }{2}-\frac{{\mathrm{a}}^2}{2}(-\frac{\pi }{2})=\frac{\pi }{2}{\mathrm{a}}^2$$$$=\frac{\pi }{2}{\left(\frac{\beta -\alpha }{2}\right)}^2=\frac{\pi }{8}{(\beta -\alpha )}^2.$$$$$$

Commented by niroj last updated on 18/May/20

it's great done..������

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