If f (x )= ( tan^2 ⁡x+ cot^2 ⁡x+ 4( sin^3 ⁡x+ cos^3 ⁡x) sin^2 ⁡2x) dx and f ( 4 )=0, then 3(f ( 6 )+2)=

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Correct answer: A. (a) f (x )= ( tan^2 ⁡x+ cot^2 ⁡x+ 4( sin^3 ⁡x+ cos^3 ⁡x) sin^2 ⁡2x) dx=( sec^2 ⁡x+ cosec^2 ⁡x-2+ tan ⁡x sec ⁡x+ cot ⁡x cosec ⁡x) dxf (x )= tan ⁡x- cot ⁡x-2x+ sec ⁡x- cosec ⁡x+Cf ( 4 )=1-1- 2+ 2- 2+C=0 ⇒C= 2 f ( 6 )= tan ⁡ 6- cot ⁡ 6+ 3+ sec ⁡ 6- cosec ⁡ 6+ 2= 1 3- 3+ 2 3-2+ 6= 6-2 Now, 3(f ( 6 )+2)= 2

$If f (x) = \int (tan^{2} x + cot^{2} x + \frac{4 ( sin ^{3} x + cos ^{3} x )}{sin ^{2} 2 x}) dx and f (\frac{π}{4}) = 0, then 3 (f (\frac{π}{6}) + 2) =$

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