If $\int \frac{\sin x}{\sin ^{3} x+\cos ^{3} x} d x=\alpha \log _{e}|1+\tan x|+\beta \log _{e}\left|1-\tan x+\tan ^{2} x\right|+\gamma \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C$, when $C$ is constant of integration, then the value of $18\left(\alpha+\beta+\gamma^{2}\right)$ is ____.

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Indefinite Integrals

\int \frac{s i n x}{s i n ^{3} x + c o s ^{3} x} d x = α lo g_{e} ∣1 + tan x ∣ + β lo g_{e} 1 - tan x + tan^{2} x + γ tan^{- 1} (\frac{2 t a n x - 1}{3}) + C

, when

C

is constant of integration, then the value of

18 (α + β + γ^{2})

is ____.

Answer: 3

Dividing numerator and denominator by

cos^{3} x

gives

\int \frac{tan x sec ^{2} x}{tan ^{3} x + 1} d x

; with

t = tan x

this is

\int \frac{t}{( t + 1 ) ( t ^{2} - t + 1 )} d t

. Partial fractions give

\frac{- 1/3}{t + 1} + \frac{( t + 1 ) /3}{t ^{2} - t + 1}

, and integrating gives

- \frac{1}{3} ln ∣1 + tan x ∣ + \frac{1}{6} ln ∣1 - tan x + tan^{2} x ∣ + \frac{1}{3} tan^{- 1} \frac{2 tan x - 1}{3} + C

. Thus

α = - \frac{1}{3}

β = \frac{1}{6}

γ = \frac{1}{3}

, and

18 (α + β + γ^{2}) = 18 (- \frac{1}{3} + \frac{1}{6} + \frac{1}{3}) = 3

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Integer answer