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Question Number 200894 by sonukgindia last updated on 26/Nov/23

Answered by Frix last updated on 26/Nov/23

$$=4\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{dx}{1+{4\sin }^{2}x}\stackrel{t=\tan x}{=}$$$$=4\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{5t^2+1}=\frac{4}{\sqrt{5}}{\left[{\tan }^{-1}\sqrt{5}t\right]}_0^{\mathrm{\infty }}=\frac{2\pi }{\sqrt{5}}$$

Answered by Calculusboy last updated on 28/Nov/23

$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:\mathit{m}\mathit{u}\mathit{l}\mathit{t}\mathit{i}\mathit{p}\mathit{l}\mathit{y}\mathit{n}\mathit{u}\mathit{m}\mathit{e}\mathit{r}\mathit{a}\mathit{t}\mathit{o}\mathit{r}\mathit{a}\mathit{n}\mathit{d}\mathit{d}\mathit{e}\mathit{n}\mathit{o}\mathit{m}\mathit{i}\mathit{n}\mathit{a}\mathit{t}\mathit{o}\mathit{r}$$$$\mathit{b}\mathit{y}\left(\frac{1}{{\mathit{c}\mathit{o}\mathit{s}}^2\mathit{x}}\right)$$$$\mathit{I}={\int }_0^{\frac{\mathit{\pi }}{2}}\frac{\frac{1}{{\mathit{c}\mathit{o}\mathit{s}}^2\mathit{x}}}{\frac{1}{{\mathit{c}\mathit{o}\mathit{s}}^2\mathit{x}+\frac{4{\mathit{s}\mathit{i}\mathit{n}}^2\mathit{x}}{{\mathit{c}\mathit{o}\mathit{s}}^2\mathit{x}}}}\mathit{d}\mathit{x}\mathit{N}\mathit{b}:\mathit{t}\mathit{a}\mathit{n}\mathit{x}=\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{x}}{\mathit{c}\mathit{o}\mathit{s}\mathit{x}}\mathit{a}\mathit{n}\mathit{d}\mathit{s}\mathit{e}\mathit{c}\mathit{x}=\frac{1}{\mathit{c}\mathit{o}\mathit{s}\mathit{x}}$$$$\mathit{I}={\int }_0^{\frac{\pi }{2}}\frac{{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{x}}{{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{x}+4{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{x}}\mathit{d}\mathit{x}\mathit{r}\mathit{e}\mathit{c}\mathit{a}\mathit{l}\mathit{l}\mathit{t}\mathit{h}\mathit{a}\mathit{t}{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{x}=1+{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{x}$$$$\mathit{I}={\int }_0^{\frac{\mathit{\pi }}{2}}\frac{{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{x}}{1+{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{x}+4{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{x}}\mathit{d}\mathit{x}\mathit{l}\mathit{e}\mathit{t}\mathit{u}=\mathit{t}\mathit{a}\mathit{n}\mathit{x}\mathit{d}\mathit{x}=\frac{\mathit{d}\mathit{u}}{{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{x}}$$$$\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathit{x}=\frac{\mathit{\pi }}{2}\mathit{u}=\mathrm{\infty }\mathit{a}\mathit{n}\mathit{d}\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathit{x}=0\mathit{u}=0$$$$\mathit{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{x}}{1+5{\mathit{u}}^2}\cdot \frac{\mathit{d}\mathit{u}}{{\mathit{s}\mathit{e}\mathit{c}}^2\mathit{x}}\iff \mathit{I}=\frac{1}{5}{\int }_0^{\mathrm{\infty }}\frac{\mathit{d}\mathit{u}}{\frac{1}{5}+{\mathit{u}}^2}$$$$\mathit{I}=\frac{1}{5}{\int }_0^{\mathrm{\infty }}\frac{\mathit{d}\mathit{u}}{{\mathit{u}}^2+{\left(\frac{1}{\sqrt{5}}\right)}^2}\iff \mathit{I}=\frac{1}{5}\times \frac{1}{\left(\frac{1}{\sqrt{5}}\right)}{\left[{\mathit{t}\mathit{a}\mathit{n}}^{-1}\left(\frac{\mathit{x}}{\left(\frac{1}{\sqrt{5}}\right)}\right)\right]}_0^{\mathrm{\infty }}+\mathit{C}$$$$\mathit{I}=\frac{\sqrt{5}}{5}[{\mathit{t}\mathit{a}\mathit{n}}^{-1}\left(\mathrm{\infty }\right)-{\mathit{t}\mathit{a}\mathit{n}}^{-1}\left(0\right)]$$$$\mathit{I}=\frac{\sqrt{5}}{5}[\frac{\mathit{\pi }}{2}-0]$$$$\mathit{I}=\frac{\mathit{\pi }\sqrt{5}}{10}$$$$$$

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