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Question Number 14630 by Don sai last updated on 03/Jun/17

$$\mathrm{solve}\mathrm{the}\mathrm{eqn}$$$$\mathrm{dr}/\mathrm{d}\theta =[\mathrm{r}({\mathrm{a}}^2-{\mathrm{r}}^2)/{\mathrm{a}}^2+{\mathrm{r}}^2]\cot \theta$$$$\mathrm{hint}.\mathrm{let}{\mathrm{a}}^2+{\mathrm{r}}^2={\mathrm{a}}^2-{\mathrm{r}}^2+{2\mathrm{r}}^2.$$

Commented by prakash jain last updated on 03/Jun/17

$$\frac{dr}{d\theta }=\frac{r(a^2-r^2)}{(a^2+r^2)}\cot \theta$$$$\frac{(a^2+r^2)}{r(a^2-r^2)}dr=\cot \theta d\theta$$$$\int \frac{(a^2+r^2)}{r(a^2-r^2)}dr$$$$(usinghint{a}^2+r^2=a^2-r^2+2r^2$$$$=\int \frac{(a^2-r^2+2r^2)}{r(a^2-r^2)}dr$$$$=\int \frac{(a^2-r^2)}{r(a^2-r^2)}dr+\int \frac{2r^2}{r(a^2-r^2)}dr$$$$=\int \frac{dr}{r}+\int \frac{2r}{(a^2-r^2)}dr$$$$sunstitute{a}^2-r^2=uinsecond$$$$part-2rdr=du$$$$=\ln r+\int \frac{-du}{u}+c$$$$=\ln r-\ln u=ln\frac{r}{a^2-r^2}+C$$$$\int \cot \theta d\theta =\ln \sin \theta +c$$$$\ln \frac{r}{a^2-r^2}=\ln \sin \theta +c$$$$\frac{r}{(a^2-r^2)\sin \theta }=e^c$$

Commented by Don sai last updated on 06/Jun/17

$$\mathrm{Thank}\mathrm{you}$$

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