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Question Number 97683 by Don08q last updated on 09/Jun/20

$$\mathrm{Evaluate}{\int }_0^1\frac{1}{\sqrt{16+9x^2}}dx$$

Commented by bemath last updated on 09/Jun/20

$$\mathrm{set}3\mathrm{x}=4\tan \mathrm{z}\Rightarrow 3\mathrm{dx}=4\sec {}^2\mathrm{zdz}$$$$\mathrm{I}=\int \frac{\frac{4}{3}\sec {}^2\mathrm{z}\mathrm{dz}}{\sqrt{16+16\tan {}^2\mathrm{z}}}$$$$\mathrm{I}=\frac{1}{3}\int \sec \mathrm{z}\mathrm{dz}=\frac{1}{3}\ln \mid \sec \mathrm{z}+\tan \mathrm{z}\mid$$$$\mathrm{I}=\frac{1}{3}{[\ln \mid \frac{\sqrt{16+{9\mathrm{x}}^2}}{4}+\frac{3\mathrm{x}}{4}\mid ]}_0^1$$$$=\frac{1}{3}\ln \left(2\right)$$

Commented by Tony Lin last updated on 09/Jun/20

$$\frac{1}{3}{\int }_0^1\frac{1}{\sqrt{{\left(\frac{4}{3}\right)}^2+x^2}}dx$$$$=\frac{1}{3}{sinh}^{-1}\left(\frac{3x}{4}\right){\mid }_0^1$$$$=\frac{1}{3}{sinh}^{-1}\left(\frac{3}{4}\right)$$

Commented by john santu last updated on 09/Jun/20

$$\mathrm{joosss}\mathrm{coolll}$$

Answered by smridha last updated on 10/Jun/20

$$\frac{1}{3}{\int }_0^1\frac{\mathit{d}\mathit{x}}{\sqrt{{\left(\frac{4}{3}\right)}^2+{\mathit{x}}^2}}=\frac{1}{3}{\left(\mathit{l}\mathit{n}[\mathit{x}+\sqrt{{\left(\frac{4}{3}\right)}^2+{\mathit{x}}^2}]\right)}_0^1$$$$=\frac{1}{3}\mathit{l}\mathit{n}2$$

Answered by Rio Michael last updated on 09/Jun/20

$$\mathcal{R}\mathrm{eca}\mathcal{L}l\int \frac{1}{\sqrt{x^2+a^2}}dx=\ln (x+\sqrt{x^2+a^2})+c$$$$\underset{0}{\stackrel{1}{\int }}\frac{1}{\sqrt{16+9x^2}}dx=\frac{1}{3}\underset{0}{\stackrel{1}{\int }}\frac{1}{\sqrt{x^2+{\left(\frac{4}{3}\right)}^2}}dx=\frac{1}{3}{\left[\ln (x+\sqrt{x^2+{\left(\frac{4}{3}\right)}^2})\right]}_0^1$$$$=\frac{1}{3}[\ln \left(\frac{8}{5}\right)-\ln \left(\frac{4}{3}\right)]=\frac{1}{3}\ln 2$$

Commented by peter frank last updated on 09/Jun/20

$$\mathrm{good}$$

Answered by mathmax by abdo last updated on 09/Jun/20

$$\mathrm{I}={\int }_0^1\frac{\mathrm{dx}}{\sqrt{16+{9\mathrm{x}}^2}}\Rightarrow \mathrm{I}=\frac{1}{4}{\int }_0^1\frac{\mathrm{dx}}{\sqrt{1+\frac{9}{16}{\mathrm{x}}^2}}\mathrm{changement}\frac{3}{4}\mathrm{x}=\mathrm{t}\mathrm{give}$$$$\mathrm{I}=\frac{1}{4}\times {\int }_0^{\frac{3}{4}}\frac{1}{\sqrt{1+{\mathrm{t}}^2}}\times \frac{4}{3}\mathrm{dt}=\frac{1}{3}[\ln {(\mathrm{t}+\sqrt{1+{\mathrm{t}}^2}]}_0^{\frac{3}{4}}=\frac{1}{3}\{\ln (\frac{3}{4}+\sqrt{1+\frac{9}{16}})$$$$=\frac{1}{3}\ln (\frac{3}{4}+\frac{5}{4})=\frac{1}{3}\ln \left(2\right)$$

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