If ∫ sin x sin 3 x + cos 3 x d x = α log e | 1 + tan x | + β log e 1 - tan x + tan 2 x + γ tan - 1 2 tan x - 1 3 + C , when C is constant of integration, then the value of 18 α + β + γ 2 is

Step-by-step solution

∫ sin x sin 3 x + cos 3 x   d x = α log e | 1 + tan x | + β log e 1 - tan x + tan 2 x + γ tan - 1 2 tan x - 1 3 + C , . . . . . . . . . . i Let, I = ∫ sin x sin 3 x + cos 3 x d x I = ∫ tan x sec 2 x tan 3 + 1 d x Put tan x = t ⇒ sec 2 x d x = d t = ∫ t d t t 3 + 1 = ∫ t ( t + 1 ) t 2 - t + 1 d t Now t ( t + 1 ) t 2 - t + 1 = A t + 1 + B t + C t 2 - t + 1 t = A t 2 - t + 1 + ( B t + C ) ( t + 1 ) t = t 2 A + B - t A - B - C + A + C Comparing on both sides A + B