Mathematics · 2019
JEE Main · 11 January 2019, Shift 2 · Q75
The integral ∫_π / 6^π / 4 (d x)/(sin 2 x( tan^5 x+ cot^5 x)) equals:
The integral $\displaystyle \int_{\pi / 6}^{\pi / 4} \frac{\mathrm{~d} x}{\sin 2 x\left(\tan ^5 x+\cot ^5 x\right)}$
equals :
Official answer
(1)
$\displaystyle \frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{9 \sqrt{3}}\right)\right)$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.