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Question Number 164600 by mathlove last updated on 19/Jan/22

Answered by Zaynal last updated on 19/Jan/22

$$1.\frac{1}{5}\mathrm{arcsec}\left(\frac{2}{5}\mathit{x}\right)+\mathbf{C}$$$$2.\frac{\mathrm{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\frac{{\mathit{t}}^2}{8}\right)}{16}+\mathit{C},\mathit{C}\in \mathbb{R}$$$$3.\frac{\mathbf{arcsin}\left(\frac{7\mathbf{y}}{3}\right)}{7}+\mathbf{C},\mathrm{C}\in \mathbb{R}$$

Answered by Mathspace last updated on 20/Jan/22

$$1)I=\int \frac{dx}{x\sqrt{4x^2-25}}$$$$\Rightarrow I{=}_{2x=5t}\frac{5}{2}\int \frac{dt}{\frac{5}{2}t.5\sqrt{t^2-1}}$$$$=\int \frac{dt}{t\sqrt{t^2-1}}{=}_{t=chu}\int \frac{shu}{chushu}du$$$$=\int \frac{du}{chu}=2\int \frac{du}{e^u+e^{-u}}$$$${=}_{e^u=y}2\int \frac{dy}{y(y+y^{-1})}=2\int \frac{dy}{y^2+1}$$$$=2arctany+C$$$$=2arctan\left(e^u\right)+C$$$$u=argcht=ln(t+\sqrt{t^2-1})\Rightarrow$$$$e^u=t+\sqrt{t^2-1}=\frac{2x}{5}+\sqrt{{\left(\frac{2x}{5}\right)}^2-1}\Rightarrow$$$$I=2arctan(\frac{2x}{5}+\sqrt{{\left(\frac{2x}{5}\right)}^2-1})+C$$

Answered by Eulerian last updated on 20/Jan/22

$$$$$$\mathbf{Solution}:$$$$1.\int \frac{\frac{1}{{\mathrm{x}}^2}}{\sqrt{4-{\left(\frac{5}{\mathrm{x}}\right)}^2}}\mathrm{dx}$$$$$$$$\stackrel{\frac{5}{\mathrm{x}}=2\mathbf{sin}\left(\mathit{\theta }\right)}{=}-\frac{2}{5}\cdot \int \frac{\cos \left(\theta \right)}{\sqrt{4-{4\sin }^2\left(\theta \right)}}\mathrm{d}\theta$$$$$$$$=-\frac{2}{5}\cdot \int \frac{1}{2}\mathrm{d}\theta$$$$=-\frac{1}{5}\cdot \theta +\mathrm{C}$$$$$$$$\mathrm{By}\mathrm{substituting}\mathrm{back}:$$$$1.-\frac{1}{5}\cdot {\sin }^{-1}\left(\frac{5}{2\mathrm{x}}\right)+\mathrm{C}$$$$$$$$$$

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