Question

If $I=\int \frac{\cos x}{{\sin }^{3}x-{\cos }^{3}x}dx$, and $f\left(x\right)=\frac{1}{3}\log (\tan x-1)+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x+1}{\sqrt{3}}\right)g\left(x\right)=\frac{1}{6}\log ({\tan }^{2}x+\tan x+1)$ then I equals

• A. $f\left(x\right)-g\left(x\right)+C$
• B. $f\left(x\right)g\left(x\right)+C$
• C. $f\left(x\right)/g\left(x\right)+C$
• D. $f\left(x\right)+g\left(x\right)+C$

Answer 1

The correct option is A $f\left(x\right)-g\left(x\right)+C$

$I=\int \frac{{\sec }^{2}x}{{\tan }^{3}x-1}dx$
Put $x=t$, so
$I=\int \frac{dt}{t^3-1}=\frac{1}{3}\int (\frac{1}{t-1}-\frac{t-1}{t^2+t+1})dt=\frac{1}{3}\log |t-1|-\frac{1}{3}{I}_1$
Where
$I_1=\int \frac{t-1}{{(t+\frac{1}{2})}^2+\frac{3}{4}}dt$
Put $t+\frac{1}{2}=u$
$I_1=\int \frac{u-\frac{3}{2}}{u^2+\frac{3}{4}}du=\frac{1}{2}\log (u^2+\frac{3}{4})-\frac{3}{2}.\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2u}{\sqrt{3}}\right)$
Hence
$I=f\left(x\right)-g\left(x\right)+C$