Question

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 ____

Answer 1

$=\int \frac{\frac{\sin x}{\cos ^{3} x}}{1+\tan ^{3} x} d x=\int \frac{\tan x \cdot \sec ^{2} x}{(\tan x+1)\left(1+\tan ^{2} x-\tan x\right)} d x$
Let $\tan x=t \Rightarrow \sec ^{2} x \cdot d x=d t$
$=\int \frac{t}{(t+1)\left(t^{2}-t+1\right)} d t$
$=\int\left(\frac{A}{t+1}+\frac{B(2 t-1)}{t^{2}-t+1}+\frac{C}{t^{2}-t+1}\right) d x$
$\Rightarrow A\left(t^{2}-t+1\right)+B(2 t-1)\left(t^{2}-t+1\right)+C(t+1)=t$
$\Rightarrow t^{2}(A+2 B)+t(-A+B+C)+A-B+C=1$
$\therefore A +2 B =0 ...$ (1)
$- A + B + C =1 ... $(2)
$A - B + C =0 ....$(3)
$\Rightarrow C =\frac{1}{2} $
$\Rightarrow A - B =-\frac{1}{2} ....$(4)
$A +2 B =0$
$A - B =-\frac{1}{2}$
$\Rightarrow 3 B =\frac{1}{2} \Rightarrow B =\frac{1}{6}$
$A =-\frac{1}{3}$
$I =-\frac{1}{3} \int \frac{ dt }{1+ t }+\frac{1}{6} \int \frac{2 t -1}{ t ^{2}- t +1} dt +\frac{1}{2} \int \frac{ dt }{ t ^{2}- t +1}$
$=-\frac{1}{3} \ell n |(1+\tan x )|+\frac{1}{6} \ln \left|\tan ^{2} x -\tan x +1\right|$
$+\frac{1}{2} \cdot \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\left(\tan x-\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}}\right)$
$=-\frac{1}{3} \ln |(1+\tan x)|+\frac{1}{6} \ln \left|\tan ^{2} x-\tan x+1\right|$
$+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C$
$\alpha=-\frac{1}{3}, \beta=\frac{1}{6}, \gamma=\frac{1}{\sqrt{3}}$
$18\left(\alpha+\beta+\gamma^{2}\right)=18\left(-\frac{1}{3}+\frac{1}{6}+\frac{1}{3}\right)=3$