The indefinite integral I = ∫ sec 2 ⁡ x tan ⁡ x sec ⁡ x + tan ⁡ x d x sec 5 ⁡ x + sec 2 ⁡ x tan 3 ⁡ x - sec 3 ⁡ x tan 2 ⁡ x - tan 5 ⁡ x simplifies to 1 3 ln ⁡ f x + c , where f π 4 = 2 2 + 1 and c is the constant of integration. If the value of f π 3 is a + b , then the value of b - 3 a is equal to

Step-by-step solution

I = 1 3 ∫ 3 sec 3 ⁡ x tan ⁡ x + 3 sec 2 ⁡ x tan 2 ⁡ x d x sec 2 ⁡ x - tan 2 ⁡ x sec 3 ⁡ x + tan 3 ⁡ x Let sec 3 ⁡ x + tan 3 ⁡ x = t Then, I = 1 3 ∫ d t t = 1 3 ln ⁡ t + c = 1 3 ln ⁡ sec 3 ⁡ x + tan 3 ⁡ x + c ∴ f x = sec 3 ⁡ x + tan 3 ⁡ x ⇒ f π 3 = 8 + 27 ⇒ a = 8 ,   b = 27 ⇒ b - 3 a = 3