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Question Number 85153 by reprins last updated on 19/Mar/20

$$\int \sqrt{4{\mathbf{x}}^2-4}\mathbf{dx}=...$$$$$$$$$$

Commented by mathmax by abdo last updated on 19/Mar/20

$$A=\int \sqrt{4x^2-4}dx=2\int \sqrt{x^2-1}dxwedothechangementx=ch\left(t\right)$$$$\Rightarrow A=2\int sh\left(t\right)sh\left(t\right)dt=2\int \frac{ch\left(2t\right)-1}{2}dt$$$$=\frac{1}{2}sh\left(2t\right)-t+C=sh\left(t\right)ch\left(t\right)-t+C$$$$=x\sqrt{x^2-1}-argch\left(x\right)+C=x\sqrt{x^2-1}-ln(x+\sqrt{x^2-1})+C$$

Answered by john santu last updated on 19/Mar/20

$$2\int \sqrt{{\mathrm{x}}^2-1}\mathrm{dx}=\mathrm{K}$$$$\mathrm{let}\mathrm{x}=\sec \mathrm{t}$$$$\mathrm{K}=2\int \sqrt{\sec {}^2\mathrm{t}-1}\sec \mathrm{t}\tan \mathrm{t}\mathrm{dt}$$$$\mathrm{K}=2\int \sec \mathrm{t}\tan {}^2\mathrm{t}\mathrm{dt}$$$$\mathrm{K}=2[\int \sec {}^3\mathrm{t}\mathrm{dt}-\ln \mid \sec \mathrm{t}+\tan \mathrm{t}\mid ]$$$$\mathrm{for}\mathrm{I}=\int \sec {}^3\mathrm{t}\mathrm{dt}=\int \sec \mathrm{t}\mathrm{d}\left(\tan \mathrm{t}\right)$$$$\mathrm{I}=\sec \mathrm{t}\tan \mathrm{t}-\int \tan {}^2\mathrm{t}\sec \mathrm{t}\mathrm{dt}$$$$\mathrm{I}=\sec \mathrm{t}\tan \mathrm{t}-\int \sec {}^3\mathrm{t}\mathrm{dt}+\ln \mid \sec \mathrm{t}+\tan \mathrm{t}\mid$$$$2\mathrm{I}=\sec \mathrm{t}\tan \mathrm{t}+\ln \mid \sec \mathrm{t}+\tan \mathrm{t}\mid$$$$\mathrm{I}=\frac{1}{2}\sec \mathrm{t}\tan \mathrm{t}+\frac{1}{2}\ln \mid \sec \mathrm{t}+\tan \mathrm{t}\mid$$$$\mathrm{K}=\sec \mathrm{t}\tan \mathrm{t}-\ln \mid \sec \mathrm{t}+\tan \mathrm{t}\mid +\mathrm{c}$$$$\therefore \mathrm{K}=\mathrm{x}\sqrt{{\mathrm{x}}^2-1}-\ln \mid \mathrm{x}+\sqrt{{\mathrm{x}}^2-1}\mid +\mathrm{c}$$

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