$$ = \int {{{{{\sin x} \over {{{\cos }^3}x}}} \over {1 + {{\tan }^3}x}}dx = \int {{{\tan x.{{\sec }^2}x} \over {(\tan x + 1)(1 + {{\tan }^2}x - \tan x)}}dx} } $$

Let $$\tan x = t \Rightarrow {\sec ^2}x.\,dx = dt$$

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

$$ = \int {\left( {{A \over {t + 1}} + {{B(2t - 1)} \over {{t^2} - t + 1}} + {C \over {{t^2} - t + 1}}} \right)dx} $$

$$ \Rightarrow A({t^2} - t + 1) + B(2t - 1)({t^2} - t + 1) + C(t + 1) = t$$

$$ \Rightarrow {t^2}(A + 2B) + t( - A + B + C) + A - B + C = 1$$

$$\therefore$$ $$A + 2B = 0$$ ..... (1)

$$ - A + B + C = 1$$ ... (2)

$$A - B + C = 0$$ ... (3)

$$ \Rightarrow C = {1 \over 2} \Rightarrow A - B = - {1 \over 2}$$ ... (4)

$$A + 2B = 0$$

$$A - B = - {1 \over 2}$$

$$\Rightarrow$$ $$3B = {1 \over 2} \Rightarrow B = {1 \over 6}$$

$$A = - {1 \over 3}$$

$$I = - {1 \over 3}\int {{{dt} \over {1 + t}} + {1 \over 6}\int {{{2t - 1} \over {{t^2} - t + 1}}dt + {1 \over 2}\int {{{dt} \over {{t^2} - t + 1}}} } } $$

$$ = - {1 \over 3}\ln |(1 + \tan x)| + {1 \over 6}\ln |{\tan ^2}x - \tan x + 1| + {1 \over 2}.{2 \over {\sqrt 3 }}{\tan ^{ - 1}}\left( {{{\left( {\tan x - {1 \over 2}} \right)} \over {{{\sqrt 3 } \over 2}}}} \right)$$

$$ = - {1 \over 3}\ln |(1 + \tan x)| + {1 \over 6}\ln |{\tan ^2}x - \tan x + 1| + {1 \over {\sqrt 3 }}{\tan ^{ - 1}}\left( {{{2\tan x - 1} \over {\sqrt 3 }}} \right) + C$$

$$\alpha = - {1 \over 3},\beta = {1 \over 6},\gamma = {1 \over {\sqrt 3 }}$$

$$18(\alpha + \beta + {\gamma ^2}) = 18\left( { - {1 \over 3} + {1 \over 6} + {1 \over 3}} \right) = 3$$