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…

$\int \frac{\sin x}{{\sin }^{3}x+{\cos }^{3}x} \mathrm{d}x=\alpha {\log }_{\mathrm{e}}|1+\tan x|+\beta {\log }_{\mathrm{e}}\left|1-\tan x+{\tan }^{2}x\right|+\gamma {\tan }^{-1}\left(\frac{2\tan x-1}{\sqrt{3}}\right)+C,..........\left(i\right)$

Let, $I=\int \frac{\sin x}{{\sin }^{3}x+{\cos }^{3}x}dx$
$I=\int \frac{\tan x{\sec }^{2}x}{{\tan }^3+1}dx$

Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

$=\int \frac{tdt}{t^3+1}=\int \frac{t}{(t+1)\left(t^2-t+1\right)}dt$
Now $\frac{t}{(t+1)\left(t^2-t+1\right)}=\frac{A}{t+1}+\frac{Bt+C}{t^2-t+1}$

$t=A\left(t^2-t+1\right)+(Bt+C)(t+1)$

$t=t^2\left(A+B\right)-t\left(A-B-C\right)+A+C$

Comparing on both sides

$A+B=0, -A+B+C=1, A+C=0$

Solving these equations
$A=-\frac{1}{3}, B=\frac{1}{3}, C=\frac{1}{3}$
Hence $I=\int \left\{\frac{-\frac{1}{3}}{t+1}+\frac{1}{3}\left(\frac{t+1}{t^2-t+1}\right)\right\}dt$
$=-\frac{1}{3}\int \frac{1}{\left(t+1\right)}dt+\frac{1}{3}\int \frac{\frac{1}{2}\left(2t-1\right)+\frac{3}{2}}{t^2-t+1}dt$

$=-\frac{1}{3}\int \frac{1}{\left(t+1\right)}dt+\frac{1}{6}\int \frac{2t-1}{\left(t^2-t+1\right)}+\frac{1}{2}\int \frac{dt}{{\left(t-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$=-\frac{1}{3}\ln \left|t+1\right|+\frac{1}{6}\ln \left|t^2-t+1\right|+\frac{1}{2}·\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)+C$

$=-\frac{1}{3}\ln \left|\tan x+1\right|+\frac{1}{6}\ln \left|{\tan }^{2}x-\tan x+1\right|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x-1}{\sqrt{3}}\right)+C$

From equation $\left(i\right)$, Comparing from both sides
$\Rightarrow \alpha =-\frac{1}{3}, \beta =\frac{1}{6}$ and $\gamma =\frac{1}{\sqrt{3}}$
So $18\left(\alpha +\beta +{\gamma }^2\right)$

$=18\left(-\frac{1}{3}+\frac{1}{6}+\frac{1}{3}\right)$

$=3$