Question

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

Answer 1

Answer: 3

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