\(I = \int \frac{\sin x}{\sin^3x + \cos^3 x} dx\)
\( = \int \frac{\tan x \sec^2 x}{\tan^3x + 1} dx\)
\(= \int \frac{\tan x + \sec^2x}{ \tan^3x + 1}dx\)
On substituting tanx = t and sec2x dx = dt, we get
\(I = \int \frac{t}{t^3 + 1} dt\)
\(= \int \frac t{(t + 1)(t^2 - t + 1)} dt\)
\(= -\frac 13 \int \frac 1{t + 1} dt + \frac13 \int \frac{t + 1}{t^2 - t + 1} dt\)
\(= -\frac 13 \log |t + 1| + \frac 16 \int \frac{2t - 1}{t^2 - t + 1} dt + \frac 12 \int \frac 1{t^2 - t + 1} dt\)
\(= -\frac 13 \log|t + 1| + \frac 16 \log|t^2 - t + 1| + \frac 12 \int \frac 1{\left(t - \frac 12\right)^2 + \left(\frac{\sqrt 3}2\right)^2} dt\)
\(= - \frac 13 \log |t + 1| + \frac 16\log|t^2 - t + 1| + \frac 1{\sqrt 3} \tan^{-1}\left(\frac{2t - 1}{\sqrt 3}\right)\)
\(= - \frac 13 \log|\tan x + 1|+ \frac 16 \log|\tan^2 x - \tan x + 1| + \frac 1{\sqrt 3} \tan^{-1}\left(\frac{2\tan x - 1}{\sqrt 3}\right) + C\)