Evaluate-0-pi-3-tan-2-xsec-x-3-dx-

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

Evaluate

\int_{0}^{\frac{π}{3}} \tan^{2} xsec (\frac{x}{3}) dx

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Answered by MJS_new last updated on 16/Oct/20

\int \tan^{2} x \sec \frac{x}{3} d x =

[t = \csc \frac{x}{3} \to d x = - 3 \tan \frac{x}{3} \sin \frac{x}{3}]

= - 3 \int \frac{{(3 t^{2} - 4)}^{2}}{{(t - 2)}^{2} {(t - 1)}^{2} {(t + 1)}^{2} {(t + 2)}^{2}} d t =

= - \frac{4}{3} \int \frac{d t}{{(t - 2)}^{2}} + \frac{2}{9} \int \frac{d t}{t - 2} - \frac{1}{12} \int \frac{d t}{{(t - 1)}^{2}} + \frac{35}{36} \int \frac{d t}{t - 1} -

- \frac{1}{12} \int \frac{d t}{{(t + 1)}^{2}} - \frac{35}{36} \int \frac{d t}{t + 1} - \frac{4}{3} \int \frac{d t}{{(t + 2)}^{2}} - \frac{2}{9} \int \frac{d t}{t + 2} =

\dots

= \frac{t (17 t^{2} - 20)}{6 (t^{4} - 5 t^{2} + 4)} + \frac{2}{9} \ln ∣ \frac{t - 2}{t + 2} ∣ + \frac{35}{36} \ln ∣ \frac{t - 1}{t + 1} ∣

now put t = \csc \frac{x}{3}

I {don}^{'} t think we get a useable exact value .

I get \approx . 713844

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