$I=\int\left(\frac{\sin x \cdot \cos x}{\sin ^{3} x+\cos ^{3} x}\right)^{2} d x$
$I=\int\left(\frac{\sin x \cdot \cos x}{\cos ^{2} x\left(1+\tan ^{3} x\right)}\right)^{2} d x$
$=\int\left(\frac{\sin x \sec ^{2} x}{\left(1+\tan ^{3} x\right)}\right)^{2} d x$
Put $1+\tan ^{3} x=t$
$d t=3 \tan ^{2} x \sec ^{2} x d x$ or $d x$
$=\frac{d t}{3 \tan ^{2} x \sec ^{2} x}$
$\therefore I=\int \frac{\sin ^{2} x \sec ^{4} x}{t^{2}} \times \frac{d t}{3 \tan ^{2} x \sec ^{2} x}$
$I = \frac{1}{3}\int {\frac{{{{\sin }^2}x \cdot {{\sec }^4}x}}{{{t^2}}}} \cdot \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} \times \frac{{dt}}{{{{\sec }^2}x}}$
$=\frac{1}{3} \int \frac{\sin ^{2} \pi \cdot \sec ^{4} x}{t^{2}} \times \frac{d t}{\sin ^{2} x \sec ^{4} x}$
$\therefore \mathrm{I}=\frac{1}{3} \int \frac{d t}{t^{2}}=\frac{1}{3} \int t^{-2} d t$
$I=\frac{1}{3}\left[\frac{t^{-2+1}}{-2+1}\right]+c$
$=\frac{-1}{3}\left[\frac{1}{t}\right]+c$
$\text { or } \mathrm{I}=-\frac{1}{3\left(1+\tan ^{3} x\right)}+c$
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