The integral equals:

Engineering

Mathematics

Properties of Definite Integral

Question

The integral $\int_{π / 6}^{π / 4} \frac{dx}{\sin 2 x (\tan^{5} x + \cot^{5} x)}$ equals:

$\frac{1}{5} (\frac{π}{4} - \tan^{- 1} (\frac{1}{3 \sqrt{3}}))$

$\frac{π}{40}$

$\frac{1}{10} (\frac{π}{4} - \tan^{- 1} (\frac{1}{9 \sqrt{3}}))$

$\frac{1}{20} \tan^{- 1} (\frac{1}{9 \sqrt{3}})$

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$I = \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{dx}{2 sinxcosx (\tan^{5} x + \cot^{5} x)}$

$= \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{\sec^{2} {xtan}^{4} x dx}{2 (\tan^{10} x + 1)}$

Put tan⁵ x = t

$\Rightarrow I = \frac{1}{10} \int_{\frac{1}{9 \sqrt{3}}}^{1} \frac{dt}{t^{2} + 1} = \frac{1}{10} (\tan^{- 1} 1 - \tan^{- 1} \frac{1}{9 \sqrt{3}})$