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Question Number 117511 by Canovas last updated on 12/Oct/20

Answered by bemath last updated on 12/Oct/20

$$\int \frac{\mathrm{dx}}{\cos {}^2\mathrm{x}(4-9\tan {}^2\mathrm{x})}=\int \frac{\sec {}^2\mathrm{x}}{4-9\tan {}^2\mathrm{x}}\mathrm{dx}$$$$=\int \frac{\mathrm{d}\left(\tan \mathrm{x}\right)}{4-9\tan {}^2\mathrm{x}}=\int \frac{\mathrm{d}\phi }{4-9{\phi }^2}$$$$=\int \frac{\mathrm{d}\phi }{(2+3\phi )(2-3\phi )}=\frac{1}{4}\int \frac{1}{2-3\phi }+\frac{1}{2+3\phi }\mathrm{d}\phi$$$$=\frac{1}{4}\{-\frac{1}{3}\ln (2-3\phi )+\frac{1}{3}\ln (2+3\phi )\}+\mathrm{c}$$$$=\frac{1}{12}\ln \mid \frac{2+3\phi }{2-3\phi }\mid +\mathrm{c}$$$$=\frac{1}{12}\ln \mid \frac{2+3\tan \mathrm{x}}{2-3\tan \mathrm{x}}\mid +\mathrm{c}$$

Commented by Canovas last updated on 12/Oct/20

$$ReallyI{haven}^{\prime }tunderstoodsir$$$$$$

Commented by bobhans last updated on 12/Oct/20

$$\mathrm{it}\mathrm{is}\mathrm{by}\mathrm{substitution}\mathrm{sir}.$$$$\phi =\tan \mathrm{x}\mathrm{so}\mathrm{d}\phi =\sec {}^2\mathrm{x}\mathrm{dx}$$

Answered by 1549442205PVT last updated on 12/Oct/20

$$\mathrm{F}=\int \frac{\mathrm{dx}}{{4\cos }^2\mathrm{x}-{9\sin }^2\mathrm{x}}=\int \frac{\frac{1}{{\cos }^2\mathrm{x}}}{4-{9\tan }^2\mathrm{x}}\mathrm{dx}$$$$=-\int \frac{1+{\tan }^2\mathrm{x}}{{9\tan }^2\mathrm{x}-4}\mathrm{dx}.\mathrm{Put}\mathrm{tanx}=\mathrm{t}$$$$\Rightarrow \mathrm{dt}=(1+{\mathrm{t}}^2)\mathrm{dx}$$$$\mathrm{F}=-\int \frac{(1+{\mathrm{t}}^2)\mathrm{dt}}{(1+{\mathrm{t}}^2)({9\mathrm{t}}^2-4)}=+\int \frac{\mathrm{dt}}{{9\mathrm{t}}^2-4}$$$$=-\frac{1}{9}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2-\frac{4}{9}}=-\frac{1}{9}\times \frac{3}{4}\int (\frac{1}{\mathrm{t}-\frac{2}{3}}-\frac{1}{\mathrm{t}+\frac{2}{3}})\mathrm{dt}$$$$=-\frac{1}{12}\ln \mid \frac{3\mathrm{t}-2}{3\mathrm{t}+2}\mid +\mathrm{C}=\frac{1}{12}\ln \mid \frac{3\mathrm{tanx}+2}{3\mathrm{tanx}-2}\mid +\mathrm{C}$$

Answered by mathmax by abdo last updated on 12/Oct/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{{4\cos }^2\mathrm{x}-{9\sin }^2\mathrm{x}}\Rightarrow \mathrm{I}=\int \frac{\mathrm{dx}}{4\frac{1+\cos \left(2\mathrm{x}\right)}{2}-9\frac{1-\cos \left(2\mathrm{x}\right)}{2}}$$$$=\int \frac{\mathrm{dx}}{2+2\cos \left(2\mathrm{x}\right)-\frac{9}{2}+\frac{9}{2}\cos \left(2\mathrm{x}\right)}=\int \frac{2\mathrm{dx}}{4+4\cos \left(2\mathrm{x}\right)-9+9\cos \left(2\mathrm{x}\right)}$$$$=\int \frac{2\mathrm{dx}}{13\cos \left(2\mathrm{x}\right)-5}{=}_{2\mathrm{x}=\mathrm{t}}\int \frac{\mathrm{dt}}{13\mathrm{cost}-5}{=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{u}}$$$$\int \frac{2\mathrm{du}}{(1+{\mathrm{u}}^2)(13\frac{1-{\mathrm{u}}^2}{1+{\mathrm{u}}^2}-5)}=\int \frac{2\mathrm{du}}{13(1-{\mathrm{u}}^2)-5(1+{\mathrm{u}}^2)}$$$$=\int \frac{2\mathrm{du}}{13-{13\mathrm{u}}^2-5-{5\mathrm{u}}^2}=\int \frac{2\mathrm{du}}{8-{18\mathrm{u}}^2}=\int \frac{\mathrm{du}}{4-{9\mathrm{u}}^2}$$$$=\int \frac{\mathrm{du}}{(2-3\mathrm{u})(2+3\mathrm{u})}=\frac{1}{4}\int (\frac{1}{2-3\mathrm{u}}+\frac{1}{2+3\mathrm{u}})\mathrm{du}=\frac{1}{4}\ln \mid 4-{9\mathrm{u}}^2\mid +\mathrm{c}$$$$=\frac{1}{4}\ln \mid 4-{9\tan }^2\left(\mathrm{x}\right)\mid +\mathrm{c}$$

Commented by bobhans last updated on 13/Oct/20

$$\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{get}\frac{1}{4}\int (\frac{1}{2-3\mathrm{u}}+\frac{1}{2+3\mathrm{u}})\mathrm{du}$$$$=\frac{1}{4}\ln \mid 4-{9\mathrm{u}}^2\mid ?\mathrm{it}\mathrm{not}\mathrm{correct}$$

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