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Question Number 55310 by afachri last updated on 21/Feb/19

$$$$$$\int 3x\sqrt{3x^3+7}dx=....$$

Commented by MJS last updated on 22/Feb/19

$$\mathrm{where}\mathrm{does}\mathrm{this}\mathrm{come}\mathrm{from}?$$$$\mathrm{I}\mathrm{tried}\mathrm{everything}\mathrm{I}\mathrm{know}\mathrm{but}\mathrm{it}\mathrm{seems}$$$$\mathrm{impossible}...\mathrm{of}\mathrm{course}3x^2\sqrt{3x^3+7}\mathrm{would}\mathrm{be}$$$$\mathrm{easy}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Feb/19

$$3x^3=7{tan}^2\theta \to x={\left(\frac{7{tan}^2\theta }{3}\right)}^{\frac{1}{3}}$$$$9x^{2}dx=14tan\theta {sec}^2\theta d\theta$$$$dx=\frac{14tan\theta {sec}^2\theta d\theta }{9{\left(\frac{7{tan}^2\theta }{3}\right)}^{\frac{2}{3}}}d\theta$$$$\int \frac{3{\left(\frac{7{tan}^2\theta }{3}\right)}^{\frac{1}{3}}\times \sqrt{7{tan}^2\theta +7}\times 14tan\theta {sec}^2\theta d\theta }{9{\left(\frac{7{tan}^2\theta }{3}\right)}^{\frac{2}{3}}}$$$$\frac{3}{9}\int \frac{\sqrt{7}sec\theta \times 14tan\theta \times {sec}^2\theta }{{\left(\frac{7}{3}\right)}^{\frac{2}{3}}\times {\left(tan\theta \right)}^{\frac{4}{3}}}d\theta$$$$\frac{1}{3}\times 14\sqrt{7}\times {\left(\frac{3}{7}\right)}^{\frac{2}{3}}\int \frac{{sec}^3\theta }{{\left(tan\theta \right)}^{\frac{1}{3}}}d\theta$$$$$$$$I=\int \frac{{sec}^3\theta }{{\left(tan\theta \right)}^{\frac{1}{3}}}d\theta$$$$sec\theta \int {sec}^2\theta \times \frac{1}{{\left(tan\theta \right)}^{\frac{1}{3}}}d\theta -\int [\frac{d}{d\theta }\left(sec\theta \right)\int \frac{{sec}^2\theta }{{\left(tan\theta \right)}^{\frac{1}{3}}}d\theta ]d\theta$$$$sec\theta \times \frac{{\left(tan\theta \right)}^{\frac{-1}{3}+1}}{\frac{2}{3}}-\int sec\theta tan\theta \times \frac{{\left(tan\theta \right)}^{\frac{2}{3}}}{\frac{2}{3}}d\theta$$$$\frac{3}{2}sec\theta {\left(tan\theta \right)}^{\frac{2}{3}}-\frac{3}{2}\int sec\theta tan\theta {({sec}^2\theta -1)}^{\frac{1}{3}}d\theta$$$$\frac{3}{2}sec\theta {\left(tan\theta \right)}^{\frac{3}{2}}-\frac{3}{2}\int {(a^2-1)}^{\frac{1}{3}}da$$$$curfewinheadquater...thatisbrain...wait...$$$$$$$$$$

Commented by afachri last updated on 21/Feb/19

$$\mathbf{it}\mathbf{is}\mathbf{hard},{\mathbf{isn}}^{\prime }\mathbf{t}\mathbf{it}\mathbf{Sir}??$$

Commented by afachri last updated on 21/Feb/19

$$\mathbf{oh}\mathbf{sir},\mathbf{i}\mathbf{think}\mathbf{my}\mathbf{mind}\mathbf{has}\mathbf{blown}.$$$$\mathbf{hehehe}...$$

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