Official solution
Given I = \[\int {{{\cos xdx} \over {{{\sin }^3}x{{\left( {1 + {{\sin }^6}x} \right)}^{2/3}}}}}\]
Let sin x = t
\[ \Rightarrow \] cos xdx = dt
\[ \therefore \] I = \[\int {{{dt} \over {{t^3}{{\left( {1 + {t^6}} \right)}^{2/3}}}}} \]
I = \[\int {{{dt} \over {{t^7}{{\left( {{1 \over {{t^6}}} + 1} \right)}^{2/3}}}}} \]
Let \[{{1 \over {{t^6}}} + 1}\] = z3
\[ \Rightarrow \] \[ - {6 \over {{t^7}}}dt = 3{z^2}dz\]
\[ \Rightarrow \] \[{{dt} \over {{t^7}}} = {{{z^2}} \over { - 2}}dz\]
So I = \[ - \int {{{{z^2}dz} \over {2{{\left( {{z^3}} \right)}^{2/3}}}}} \]
I = \[ - \int {{{dz} \over 2}} \]
I = \[ - {z \over 2} + C\]
I = \[ - {1 \over 2}{\left( {1 + {1 \over {{t^6}}}} \right)^{{1 \over 3}}} + C\]
I = \[ - {1 \over 2}{\left( {1 + {1 \over {{{\sin }^6}x}}} \right)^{{1 \over 3}}} + C\]
I = \[ - {1 \over {2{{\sin }^2}x}}{\left( {{{\sin }^6}x + 1} \right)^{{1 \over 3}}} + C\]
\[ \therefore \] \[\lambda \] = 3 and f(x) = \[ - {1 \over 2}\]cosec2 x
Then \[\lambda f\left( {{\pi \over 3}} \right)\] = 3 \[ \times \] \[ - {1 \over 2}\]cosec2 \[{\pi \over 3}\]
= 3 \[ \times \] \[ - {1 \over 2}\] \[ \times \] \[{4 \over 3}\] = -2