Correct Answer - Option 2 : 2/3
\(\mathop \smallint \nolimits_0^{\frac{\pi }{2}} {\cos ^3}x\;dx\)
= \(\mathop \smallint \nolimits_0^{\frac{\pi }{2}} \cos {\rm{x}}({\cos ^2}x)\;dx\)
= \(\mathop \smallint \nolimits_0^{\frac{\pi }{2}} \cos x\;(1 - {\sin ^2}x)\;dx\)
= \(\mathop \smallint \nolimits_0^{\frac{\pi }{2}} (\cos x - \cos x\;{\sin ^2}x)\;dx\)
= \(\mathop \smallint \nolimits_0^{\frac{\pi }{2}} \cos x\;dx - \mathop \smallint \nolimits_0^{\frac{\pi }{2}} \cos x{\sin ^2}x\;dx\)
u = sin x, u → 0 as x → 0, u → 1 as x → π/2
\(\frac{{du}}{{dx}} = cosx,\;du = \cos x\;dx\)
= \(\left[ {\sin x} \right]_0^{\frac{\pi }{2}} - \mathop \smallint \nolimits_0^{\frac{\pi }{2}} {u^2}du\)
= \(1 - \left( {\frac{{{u^3}}}{3}} \right)_0^1\)
= \(1 - \frac{1}{3}\)
\(\left( {\frac{2}{3}} \right)\)
