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Question Number 73832 by FCB last updated on 16/Nov/19

Commented by FCB last updated on 16/Nov/19

$$\mathbf{solve}$$

Commented by MJS last updated on 16/Nov/19

$$\mathrm{welcome}\mathrm{back}\mathrm{with}\mathrm{another}\mathrm{account}!$$

Commented by peter frank last updated on 16/Nov/19

$$pleasesirmjscheckworking$$$$$$

Commented by MJS last updated on 16/Nov/19

$$\mathrm{you}\mathrm{went}\mathrm{in}\mathrm{one}\mathrm{direction}\mathrm{and}\mathrm{then}\mathrm{turned}$$$$\mathrm{around}\mathrm{and}\mathrm{now}{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{from}\mathrm{where}\mathrm{you}$$$$\mathrm{started}$$$$\sqrt{1+\frac{1}{u^4}}=\sqrt{\frac{u^4+1}{u^4}}=\frac{\sqrt{u^4+1}}{u^2}$$

Commented by peter frank last updated on 16/Nov/19

$$absolutetruesir.pleasehelpme$$

Commented by FCB last updated on 16/Nov/19

$$????$$

Commented by MJS last updated on 16/Nov/19

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{possible}\mathrm{to}\mathrm{solve}\mathrm{this}\mathrm{integral}\mathrm{using}$$$$\mathrm{basic}\mathrm{calculus}$$

Answered by peter frank last updated on 16/Nov/19

$$x^2=\tan \theta x=\sqrt{\tan \theta }$$$$dx=\frac{\sec {}^{2}d\theta }{2\sqrt{\tan \theta }}$$$$\int \frac{\sqrt{1+\tan {}^2}}{\tan \theta }dx$$$$\int \frac{\sqrt{1+\tan {}^2}}{\tan \theta }.\frac{\sec {}^{2}d\theta }{2\sqrt{\tan \theta }}$$$$\int \frac{\sec \theta }{\tan \theta }.\frac{\sec {}^{2}d\theta }{2\sqrt{\tan \theta }}$$$$\frac{1}{2}\int \frac{\sec \theta }{\tan \theta }.\frac{\sec {}^2\theta d\theta }{\sqrt{\tan \theta }}$$$$\frac{1}{2}\int \frac{\sec {}^3\theta }{\tan \theta \sqrt{\tan \theta }}d\theta$$$$u=\sqrt{\tan \theta }{u}^2=\tan \theta$$$$d\theta =\frac{2u}{\sec {}^2\theta }du$$$$\frac{1}{2}\int \frac{\sec {}^3\theta }{u^2.u}.\frac{2u}{\sec {}^2\theta }$$$$\int \frac{\sec \theta }{u^2}$$$$\int \frac{\sqrt{1+\tan {}^2\theta }}{u^2}$$$$\int \sqrt{\frac{1+u^4}{u^4}}$$$$\int \sqrt{1+\frac{1}{u^4}}$$$$........$$

Commented by FCB last updated on 16/Nov/19

Commented by FCB last updated on 16/Nov/19

$$\mathbf{prove}\mathbf{that}$$

Commented by FCB last updated on 16/Nov/19

Commented by FCB last updated on 16/Nov/19

$$\mathrm{sir}\mathrm{peter}\mathrm{frank}?$$

Commented by peter frank last updated on 16/Nov/19

$$yesitismistake.Ihave$$$$rectifiedpleasechecknow$$

Commented by mind is power last updated on 16/Nov/19

$$\int \frac{\sqrt{1+{\mathrm{x}}^4}}{{\mathrm{x}}^2}\mathrm{dx}=\frac{-\sqrt{1+{\mathrm{x}}^4}}{\mathrm{x}}+\int \frac{{2\mathrm{x}}^3}{\mathrm{x}\sqrt{1+{\mathrm{x}}^4}}\mathrm{dx}$$$$\mathrm{we}\mathrm{devloppe}$$$$\int \frac{{2\mathrm{x}}^2}{\sqrt{1+{\mathrm{x}}^4}}\mathrm{dx}$$$$\mathrm{x}={\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}.\mathrm{sh}\left(\mathrm{t}\right)\Rightarrow \mathrm{dx}={\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}\mathrm{ch}\left(\mathrm{t}\right)$$$$\Rightarrow {\int }_0^{\mathrm{x}}\frac{{2\mathrm{x}}^2\mathrm{dx}}{\sqrt{1+{\mathrm{x}}^4}}={2\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}{\int }_0^{{\mathrm{sh}}^{-}\left({\mathrm{xe}}^{-\frac{\mathrm{i}\pi }{4}}\right)}.\frac{{\mathrm{ish}}^2\left(\mathrm{t}\right)\mathrm{ch}\left(\mathrm{t}\right)}{\sqrt{1-{\mathrm{sh}}^4\left(\mathrm{t}\right)}}$$$$={2\mathrm{ie}}^{\mathrm{i}\frac{\pi }{4}}{\int }_0^{{\mathrm{sh}}^{-}\left({\mathrm{xe}}^{-\frac{\mathrm{i}\pi }{4}}\right)}\frac{{\mathrm{sh}}^2\left(\mathrm{t}\right)\mathrm{ch}\left(\mathrm{t}\right)}{\sqrt{(1-{\mathrm{sh}}^2\left(\mathrm{t}\right))(1+{\mathrm{sh}}^2\left(\mathrm{t}\right))}}\mathrm{dt}$$$$={2\mathrm{ie}}^{\mathrm{i}\frac{\pi }{4}}{\int }_0^{{\mathrm{sh}}^{-}\left({\mathrm{xe}}^{-\frac{\mathrm{i}\pi }{4}}\right)}\frac{{\mathrm{sh}}^2\left(\mathrm{t}\right)\mathrm{ch}\left(\mathrm{t}\right)}{\sqrt{1-{\mathrm{sh}}^2\left(\mathrm{t}\right)}.\mathrm{ch}\left(\mathrm{t}\right)}\mathrm{dt}$$$${2\mathrm{ie}}^{\mathrm{i}\frac{\pi }{4}}{\int }_0^{{\mathrm{sh}}^{-}\left({\mathrm{xe}}^{-\frac{\mathrm{i}\pi }{4}}\right)}\frac{{\mathrm{sh}}^2\left(\mathrm{t}\right)}{\sqrt{1-{\mathrm{sh}}^2\left(\mathrm{t}\right)}.}\mathrm{dt}$$$$\mathrm{t}=\mathrm{is}$$$$\mathrm{sh}\left(\mathrm{is}\right)=\mathrm{isin}\left(\mathrm{s}\right)$$$$\iff -{2\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}{\int }_0^{-{\mathrm{ish}}^{-}\left({\mathrm{xe}}^{-\frac{\mathrm{i}\pi }{4}}\right)}\frac{{\sin }^2\left(\mathrm{t}\right)}{\sqrt{1+{\sin }^2\left(\mathrm{t}\right)}}$$$$\iff -{2\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}{\int }_0^{-{\mathrm{ish}}^{-}\left({\mathrm{xe}}^{-\frac{\mathrm{i}\pi }{4}}\right)}\frac{{\sin }^2\left(\mathrm{t}\right)+1-1}{\sqrt{1+{\sin }^2\left(\mathrm{t}\right)}}$$$$\iff -{2\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}{\int }_0^{-{\mathrm{ish}}^{-}\left({\mathrm{xe}}^{-\frac{\mathrm{i}\pi }{4}}\right)}\sqrt{1-{\left(\mathrm{i}\right)}^2{\sin }^2\left(\mathrm{t}\right)}-2(-{\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}){\int }_0^{-{\mathrm{ish}}^{-}\left({\mathrm{xe}}^{-\frac{\mathrm{i}\pi }{4}}\right)}\frac{\mathrm{dt}}{\sqrt{1-{\left(\mathrm{i}\right)}^2{\sin }^2\left(\mathrm{t}\right)}}$$$$\mathrm{E}(\theta \rfloor {\mathrm{k}}^2)={\int }_0^{\theta }\frac{\mathrm{d}\phi }{\sqrt{1-{\mathrm{k}}^2{\sin }^2\left(\phi \right)}},2\mathrm{nd}\mathrm{eleptic}\mathrm{function}$$$$\mathrm{F}\left(\theta \right]{\mathrm{k}}^2)={\int }_0^{\theta }\sqrt{1-\left({\mathrm{k}}^2\right){\sin }^2\left(\phi \right)}\mathrm{d}\phi$$$$\mathrm{so}\mathrm{we}\mathrm{get}$$$$-{2\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}\mathrm{F}(-{\mathrm{ish}}^{-}\left({\mathrm{xe}}^{\mathrm{i}\frac{\pi }{4}}\right)\mid -1)-(-{2\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}})\mathrm{E}(-\mathrm{ish}\left({\mathrm{xe}}^{\mathrm{i}\frac{\pi }{4}}\right)\mid -1)$$$$\mathrm{to}\mathrm{finish}{\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}={(-1)}^{\frac{1}{4}}$$$$\int \frac{\sqrt{1+{\mathrm{x}}^4}}{{\mathrm{x}}^2}\mathrm{dx}=\frac{-\sqrt{1+{\mathrm{x}}^4}}{\mathrm{x}}+\int \frac{{2\mathrm{x}}^3}{\mathrm{x}\sqrt{1+{\mathrm{x}}^4}}\mathrm{dx}$$$$-\frac{\sqrt{1+{\mathrm{x}}^4}}{\mathrm{x}}=-\frac{1}{\mathrm{x}\sqrt{1+{\mathrm{x}}^4}}-\frac{{\mathrm{x}}^3}{\sqrt{1+{\mathrm{x}}^4}}$$$$\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{answer}$$$$$$$$$$

Commented by MJS last updated on 17/Nov/19

$$\mathrm{yes};\mathrm{thank}\mathrm{you}$$

Commented by mind is power last updated on 17/Nov/19

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{welcom}\mathrm{sir}$$

Commented by FCB last updated on 17/Nov/19

$$\mathrm{great}!$$

Answered by Tanmay chaudhury last updated on 17/Nov/19

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