if

Let $I=\int \frac{\cos x d x}{\sin ^{3} x\left(1+\sin ^{6} x\right)^{2 / 3}}$

$=f(x)\left(1+\sin ^{6} x\right)^{1 / \lambda}+c$ ...(i)

If $\sin x=t$

then, $\cos x d x=d t$

$I=\int \frac{d t}{t^{3}\left(1+t^{6}\right)^{\frac{2}{3}}}=\int \frac{d t}{t^{7}\left(1+\frac{1}{t^{6}}\right)^{\frac{2}{3}}}$

Put $1+\frac{1}{t^{6}}=r^{3} \Rightarrow \frac{d t}{t^{7}}=\frac{-1}{2} r^{2} d r$

$-\frac{1}{2} \int \frac{r^{2} d r}{r^{2}}=-\frac{1}{2} r+c$

$=-\frac{1}{2}\left(\frac{\sin ^{6} x+1}{\sin ^{6} x}\right)^{\frac{1}{3}}+c$

$=-\frac{1}{2 \sin ^{2} x}\left(1+\sin ^{6} x\right)^{\frac{1}{3}}+c$

$f(x)=-\frac{1}{2} \operatorname{cosec}^{2} x$ and $\lambda=3$ [from eqn. (i)]

$\therefore \lambda f\left(\frac{\pi}{3}\right)=-2$