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Question Number 141943 by cesarL last updated on 25/May/21

$$\int \frac{dx}{x\sqrt{16-4x^2}}$$

Answered by MJS_new last updated on 25/May/21

$$\int \frac{dx}{x\sqrt{16-4x^2}}=\frac{1}{2}\int \frac{dx}{x\sqrt{4-x^2}}=$$$$[t=\frac{2+\sqrt{4-x^2}}{x}\to dx=-\frac{x^2\sqrt{4-x^2}}{2(2+\sqrt{4-x^2})}dt]$$$$=-\frac{1}{4}\int \frac{dt}{t}=-\frac{1}{4}\ln t=$$$$=\frac{1}{4}\ln \mid x\mid -\frac{1}{4}\ln (2+\sqrt{4-x^2})+C$$

Commented by cesarL last updated on 25/May/21

$$Ineedwithtrigonometricsustitution$$

Commented by Ar Brandon last updated on 25/May/21

$$\mathrm{Then}\mathrm{you}\mathrm{may}\mathrm{let}\mathrm{x}=2\sin \theta$$

Answered by mathmax by abdo last updated on 25/May/21

$$\mathrm{\Psi }=\int \frac{\mathrm{dx}}{\mathrm{x}\sqrt{16-{4\mathrm{x}}^2}}\Rightarrow \mathrm{\Psi }=\int \frac{\mathrm{dx}}{2\mathrm{x}\sqrt{4-{\mathrm{x}}^2}}[\mathrm{changement}\mathrm{x}=2\mathrm{sint}\mathrm{give}$$$$\mathrm{\Psi }=\int \frac{2\mathrm{cost}}{4\mathrm{sint}.2\mathrm{cost}}\mathrm{dt}=\frac{1}{4}\int \frac{\mathrm{dt}}{\mathrm{sint}}{=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{y}}\frac{1}{4}\int \frac{2\mathrm{dy}}{(1+{\mathrm{y}}^2)\frac{2\mathrm{y}}{1+{\mathrm{y}}^2}}$$$$=\frac{1}{4}\int \frac{\mathrm{dy}}{\mathrm{y}}=\frac{1}{4}\log \mid \mathrm{y}\mid +\mathrm{C}=\frac{1}{4}\log \mid \tan \frac{\mathrm{t}}{2}\mid +\mathrm{C}$$$$\mathrm{t}=\arcsin \left(\frac{\mathrm{x}}{2}\right)\Rightarrow \mathrm{\Psi }=\frac{1}{4}\log \mid \tan (\frac{1}{2}\arcsin \left(\frac{\mathrm{x}}{2}\right)\mid +\mathrm{C}$$

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