The integral ∫π/3π/6 sec²/³x.cosec⁴/³x dx equal to:

The correct option is A 37/6−35/6
I=∫1cos²/³xsin¹/³xsinxdx=∫tan²/³xtan²x.sec²xdx=∫sec²xtan⁴/³xdx
(tanx=t;sec²xdx=dt)=∫t⁴/³dt=t⁷/³/(7/3)=3/7t⁷/³
⇒I=3/7tan⁷/³(x) ⇒I=3/7[tan⁷/³(x)]π/3π/6
=3/7[(√3)⁷/³−(1/√3)⁷/³]=3/7[3⁷/⁶−(1/3)⁷/⁶]=3/7[3⁷/⁶−1/3⁷/⁶]
=3/7[3⁷/⁶−1/3⁷/⁶]=3/7(3⁷/⁶−3⁻⁷/⁶)=3(3⁷/⁶−3⁻⁷/⁶)/7
=3(3⁷/⁶−3⁻⁷/⁶)/7 =3(3⁷/⁶-3⁻⁷/⁶)/7=3(3⁷/⁶−(1/3)⁷/⁶)/7≈3(3⁷/⁶−3⁻⁷/⁶)/7
=3(3⁷/⁶−1/3⁷/⁶)/7≈3(4.3267−0.1551)/7≈3(4.1716)/7≈1.781
37/6−35/6=2/6=1/3≈0.3333
35/3−31/3=4/3≈1.3333
34/3−31/3=3/3=1
35/6−32/3=35/6−64/6=−29/6≈−4.833
Let's recalculate:
I = ∫(π/3) to (π/6) sec^(2/3)x cosec^(4/3)x dx
= ∫(π/3) to (π/6) (1/cos^(2/3)x) (1/sin^(4/3)x) dx
=∫(π/3) to (π/6) (sin x/ cos^(2/3)x sin^(7/3)x ) dx
Let tanx =t, sec^2 x dx =dt
I = ∫ tan^(4/3)x sec^2 x dx =∫ t^(4/3) dt =(3/7) t^(7/3)
= (3/7) [tan^(7/3)x] (π/3) to (π/6) = (3/7) [ (√3)^(7/3) – (1/√3)^(7/3)]
= (3/7) [ 3^(7/6) – 3^(-7/6)] = (3/7) [ 3^(7/6) – 1/3^(7/6)]
≈ (3/7) [ 4.3267 – 0.1551] ≈ (3/7) [4.1716] ≈ 1.781
37/6 - 35/6 = 2/6 = 1/3 ≈ 0.333