Question

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

Answer 1

$I=\int \frac{\sin x}{{\cos }^{3}x+{\sin }^{3}x}dx=\int \frac{\frac{\sin x}{{\cos }^{3}x}}{\frac{{\cos }^{3}x}{{\cos }^{3}x}+\frac{{\sin }^{3}x}{{\cos }^{3}x}}dx$

$I=\int \frac{\tan x\cdot {\sec }^{2}x}{1+{\tan }^{3}x}dx$
Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$I=\int \frac{t}{1+t^3}dt$
$\frac{t}{1+t^3}=\frac{t}{(1+t)(1+t^2-t)}=\frac{A}{1+t}+\frac{Bt+C}{1+t^2-t}$
$t=A(1-t+t^2)+(1+t)(Bt+C)$
By comparing coefficient of $t,t^2$ and constant term,
$A=-\frac{1}{3},B=\frac{1}{3},C=\frac{1}{3}$
$I=-\frac{1}{3}\int \frac{1}{1+t}dt+\frac{1}{3}\int \frac{t+1}{t^2-t+1}dt=-\frac{1}{3}\ln (1+t)+\frac{1}{6}\left[\int \frac{2t-1}{t^2-t+1}dt+3\int \frac{1}{t^2-t+1}dt\right]=-\frac{1}{3}\ln (1+t)+\frac{1}{6}\left[\int \frac{2t-1}{t^2-t+1}dt+3\int \frac{1}{{(t-1/2)}^2+3/4}dt\right]=-\frac{1}{3}\ln (1+t)+\frac{1}{6}[\ln (t^2-t+1)+3\cdot \frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)]+CI=-\frac{1}{3}\ln (1+\tan x)+\frac{1}{6}\cdot \ln ({\tan }^{2}x-\tan x+1)+\frac{1}{\sqrt{3}}\cdot {\tan }^{-1}\left(\frac{2\tan x-1}{\sqrt{3}}\right)+C\alpha =-\frac{1}{3},\beta =\frac{1}{6},\gamma =\frac{1}{\sqrt{3}}$
$\therefore 18(\alpha +\beta +{\gamma }^2)$
$=18(-\frac{1}{3}+\frac{1}{6}+\frac{1}{3})=3$