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Question Number 171910 by Tawa11 last updated on 21/Jun/22

$$\int \frac{\mathrm{dx}}{9-{4\mathrm{x}}^2}$$$$\mathrm{using}\mathrm{the}\mathrm{trigonometric}\mathrm{substitution}.$$

Answered by aleks041103 last updated on 21/Jun/22

$$\int \frac{dx}{9-4x^2}=\frac{1}{9}\int \frac{dx}{1-{\left(\frac{2x}{3}\right)}^2}=\frac{1}{9}\frac{3}{2}\int \frac{du}{1-u^2}=$$$$=\frac{1}{6}\int \frac{du}{1-u^2}$$$$\frac{1}{1-u^2}=\frac{1}{(1-u)(1+u)}=\frac{1}{2}(\frac{1}{1-u}+\frac{1}{1+u})$$$$\int \frac{du}{1-u^2}=\frac{1}{2}ln\mid 1+u\mid -\frac{1}{2}ln\mid 1-u\mid$$$$\Rightarrow \int \frac{dx}{9-4x^2}=\frac{1}{12}ln\mid \frac{1+\frac{2}{3}x}{1-\frac{2}{3}x}\mid$$$$\Rightarrow \int \frac{dx}{9-4x^2}=\frac{1}{12}ln\mid \frac{3+2x}{3-2x}\mid$$

Commented by Tawa11 last updated on 21/Jun/22

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.$$

Commented by Tawa11 last updated on 21/Jun/22

$$\mathrm{But}\mathrm{sir},\mathrm{I}\mathrm{need}\mathrm{the}\mathrm{trigonometric}[\mathrm{substion}.$$$$\mathrm{like}.\mathrm{let}\mathrm{x}=3\tan \theta$$

Commented by Tawa11 last updated on 21/Jun/22

$$\mathrm{Can}\mathrm{the}\mathrm{question}\mathrm{be}\mathrm{solved}\mathrm{using}\mathrm{trigonometric}\mathrm{substitution}?$$

Commented by mr W last updated on 21/Jun/22

$$certainlyyoucan!thatmeansyou$$$$canalwaysmakeeasythingsmore$$$$complicated,whenyoulike.$$

Commented by Tawa11 last updated on 21/Jun/22

$$\mathrm{Because}\mathrm{the}\mathrm{condition}\mathrm{given}\mathrm{is}\mathrm{using}\mathrm{trigonometric}\mathrm{substitution}\mathrm{sir}.$$

Commented by mr W last updated on 21/Jun/22

$$youcandoit!ijustsaiditisnota$$$$goodidea.trywithu=\frac{3}{2}\tan \theta ,you$$$$willsee.$$

Commented by Tawa11 last updated on 21/Jun/22

$$\mathrm{Thanks}\mathrm{sir}.$$

Commented by Tawa11 last updated on 22/Jun/22

$$\int \frac{\mathrm{dx}}{9-{4\mathrm{x}}^2}$$$$=\int \frac{\frac{3}{2}{\sec }^2\theta }{9-9{\tan }^2\theta }\mathrm{d}\theta$$$$=\frac{3}{18}\int \frac{{\sec }^2\theta }{1-{\tan }^2\theta }\mathrm{d}\theta$$$${\sec }^2\theta =1+{\tan }^2\theta$$$$\mathrm{From}\mathrm{here},\mathrm{please}\mathrm{what}\mathrm{next}...$$

Answered by ajfour last updated on 22/Jun/22

$$letI=\int \frac{dx}{9-4x^2}=4\int \frac{dx}{36-16x^2}$$$$=4\int \frac{dx}{(6-4x)(6+4x)}$$$$=\frac{4}{12}\int \frac{(6-4x)+(6+4x)}{(6-4x)(6+4x)}dx$$$$=\frac{1}{3}\int \frac{dx}{6+4x}+\frac{1}{3}\int \frac{dx}{6-4x}$$$$=\frac{1}{12}\int \frac{4dx}{6+4x}-\frac{1}{12}\int \frac{-4dx}{6-4x}$$$$=\frac{1}{12}\ln \mid 6+4x\mid -\frac{1}{12}\ln \mid 6-4x\mid +c$$$$=\frac{1}{12}\ln \mid \frac{3+2x}{3-2x}\mid +c$$

Commented by Tawa11 last updated on 22/Jun/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by thfchristopher last updated on 22/Jun/22

$$\mathrm{Let}x=\frac{3}{2}\sin \theta$$$$dx=\frac{3}{2}\cos \theta d\theta$$$$\sin \theta =\frac{2x}{3},\tan \theta =\frac{2x}{\sqrt{9-4x^2}},\sec \theta =\frac{3}{\sqrt{9-4x^2}}$$$$\therefore \int \frac{dx}{9-4x^2}$$$$=\frac{3}{2}\int \frac{\cos \theta }{{9\cos }^2\theta }d\theta$$$$=\frac{1}{6}\int \sec \theta d\theta$$$$=\frac{1}{6}\ln \mid \tan \theta +\sec \theta \mid +C$$$$=\frac{1}{6}\ln \mid \frac{3+2x}{\sqrt{9-4x^2}}\mid +C$$$$=\frac{1}{6}\ln \mid \sqrt{\frac{3+2x}{3-2x}}\mid +C$$$$=\frac{1}{12}\ln \mid \frac{3+2x}{3-2x}\mid +C$$

Commented by thfchristopher last updated on 22/Jun/22

$$\mathrm{Trigonometric}\mathrm{substitution}\mathrm{applied}\mathrm{on}$$$$\mathrm{integral}\mathrm{format}:$$$$\left(\mathrm{i}\right)\int \frac{dx}{a^2+b^2{x}^2}\mathrm{or}\int \frac{dx}{b^2{x}^2+a^2}$$$$\mathrm{Let}bx=a\tan \theta$$$$\left(\mathrm{ii}\right)\int \frac{dx}{a^2-b^2{x}^2}$$$$\mathrm{Let}bx=a\sin \theta$$$$\left(\mathrm{iii}\right)\int \frac{dx}{b^2{x}^2-a^2}$$$$\mathrm{Let}bx=a\sec \theta$$

Commented by Tawa11 last updated on 22/Jun/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa11 last updated on 22/Jun/22

$$\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

Commented by mr W last updated on 22/Jun/22

$$asubstitution{doesn}^{\prime }tmakethething$$$$easierinanyform,thereforeunnecesary.$$$$e.g.$$$$\int \frac{dx}{a^2-b^2{x}^2}$$$$=\frac{1}{2a}\int (\frac{1}{a-bx}+\frac{1}{a+bx})dx$$$$=\frac{1}{2ab}[-\ln (a-bx)+\ln (a+bx)]$$$$=\frac{1}{2ab}\times \ln \mid \frac{a+bx}{a-bx}\mid +C$$$$\int \frac{dx}{b^2{x}^2-a^2}=-\int \frac{dx}{a^2-b^2{x}^2}=-\frac{1}{2ab}\times \ln \mid \frac{a+bx}{a-bx}\mid +C$$

Commented by Tawa11 last updated on 22/Jun/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

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