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Question Number 200978 by sonukgindia last updated on 27/Nov/23

Answered by BaliramKumar last updated on 27/Nov/23

$$\mathrm{put}\mathrm{x}=\tan \theta \Rightarrow \mathrm{dx}={\sec }^2\theta \mathrm{d}\theta$$$$\int \frac{1}{\sqrt{{\tan }^2\theta +1}}{\sec }^2\theta \mathrm{d}\theta$$$$\int \sec \theta \mathrm{d}\theta =\ln (\sec \theta +\tan \theta )$$$$\Rightarrow \ln (\sqrt{{\tan }^2\theta +1}+\tan \theta )$$$$\Rightarrow {\left[\ln (\sqrt{{\mathrm{x}}^2+1}+\mathrm{x})\right]}_0^{\frac{1}{2}}=\ln \left(\frac{1+\sqrt{5}}{2}\right)=0.4812$$$$$$$$$$

Answered by MM42 last updated on 27/Nov/23

$${I=ln(x+\sqrt{1+x^2})]}_0^{\frac{1}{2}}=ln\left(\frac{1+\sqrt{5}}{2}\right)✓$$$$$$

Answered by Calculusboy last updated on 28/Nov/23

$$\mathit{R}\mathit{e}\mathit{c}\mathit{a}\mathit{l}\mathit{l}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\int \frac{\mathit{d}\mathit{x}}{\sqrt{{\mathit{x}}^2+{\mathit{a}}^2}}={\mathit{s}\mathit{i}\mathit{n}\mathit{h}}^{-1}\left(\frac{\mathit{x}}{\mathit{a}}\right)+\mathit{C}$$$$\mathit{t}\mathit{h}\mathit{e}\mathit{n},$$$${\int }_0^{\frac{1}{2}}\frac{\mathit{d}\mathit{x}}{\sqrt{{\mathit{x}}^2+1}}=\mid {\mathit{s}\mathit{i}\mathit{n}\mathit{h}}^{-1}\left(\frac{\mathit{x}}{1}\right){\mid }_0^{\frac{1}{2}}+C$$$$\Rightarrow {\mathit{s}\mathit{i}\mathit{n}\mathit{h}}^{-1}\left(\frac{1}{2}\right)-{\mathit{s}\mathit{i}\mathit{n}\mathit{h}}^{-1}\left(0\right)=0.4812(4.\mathit{d}.\mathit{p})$$

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