Question

Find: $\int \frac{\sin x}{{\sin }^{3}x+{\cos }^{3}x}dx$.

Answer 1

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

$=\int \frac{\tan x{\sec }^{2}x}{{\tan }^{3}x+1}$

On substituting $\tan x=t$ and ${\sec }^{2}xdx=dt,$ we get

$I=\int \frac{t}{t^3+1}dt$

$=\int \frac{t}{(t+1)(t^2-t+1)}dt$

$=\frac{-1}{3}\int \frac{dt}{t+1}+\frac{1}{3}\int \frac{t+1}{t^2-t+1}dt$

$=\frac{-1}{3}\log |t+1|+\frac{1}{6}\int \frac{(2t-1)+3}{t^2-t+1}dt$

$=\frac{-1}{3}\log |t+1|+\frac{1}{6}\int \frac{(2t-1)}{t^2-t+1}dt+\frac{1}{2}\int \frac{1}{t^2-t+1}dt$

$=\frac{-1}{3}\log |t+1|+\frac{1}{6}\log |t^2-t+1|+\frac{1}{2}\int \frac{dt}{{(t-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$=\frac{-1}{3}\log |t+1|+\frac{1}{6}\log |t^2-t+1|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)$

$=\frac{-1}{3}\log |\tan x+1|+\frac{1}{6}\log |{\tan }^{2}x-\tan x+1|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x-1}{\sqrt{3}}\right)$ where $t=\tan x$