The integral _ / 6 ^ / 3 ^ 2 / 3 x cosec ^ 4 / 1 x d x is equal to:
Mathematics · JEE Main · NTA Exams — Integral Calculus
The integral \[\int_{\pi / 6}^{\pi / 3} \sec ^{2 / 3} x \operatorname{cosec}^{4 / 1} x d x\] is equal to:
- \(3^{\frac{5}{3}}-3^{\frac{1}{3}}\)
- \(3^{\frac{5}{6}}-3^{\frac{2}{3}}\)
- \(3^{\frac{4}{3}}-3^{\frac{1}{3}}\)
- \(3^{\frac{7}{6}}-3^{\frac{5}{6}}\)
Answer
(A) 3^ 5 3 -3^ 1 3
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