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Question Number 118278 by Lordose last updated on 16/Oct/20

$$\mathrm{Evaluate}$$$${\int }_0^{\frac{\pi }{3}}{\tan }^2\mathrm{xsec}\left(\frac{\mathrm{x}}{3}\right)\mathrm{dx}$$$$★$$

Answered by MJS_new last updated on 16/Oct/20

$$\int {\tan }^{2}x\sec \frac{x}{3}dx=$$$$[t=\csc \frac{x}{3}\to dx=-3\tan \frac{x}{3}\sin \frac{x}{3}]$$$$=-3\int \frac{{(3t^2-4)}^2}{{(t-2)}^2{(t-1)}^2{(t+1)}^2{(t+2)}^2}dt=$$$$=-\frac{4}{3}\int \frac{dt}{{(t-2)}^2}+\frac{2}{9}\int \frac{dt}{t-2}-\frac{1}{12}\int \frac{dt}{{(t-1)}^2}+\frac{35}{36}\int \frac{dt}{t-1}-$$$$-\frac{1}{12}\int \frac{dt}{{(t+1)}^2}-\frac{35}{36}\int \frac{dt}{t+1}-\frac{4}{3}\int \frac{dt}{{(t+2)}^2}-\frac{2}{9}\int \frac{dt}{t+2}=$$$$...$$$$=\frac{t(17t^2-20)}{6(t^4-5t^2+4)}+\frac{2}{9}\ln \mid \frac{t-2}{t+2}\mid +\frac{35}{36}\ln \mid \frac{t-1}{t+1}\mid$$$$\mathrm{now}\mathrm{put}t=\csc \frac{x}{3}$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{we}\mathrm{get}\mathrm{a}\mathrm{useable}\mathrm{exact}\mathrm{value}.$$$$\mathrm{I}\mathrm{get}\approx .713844$$

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