Correct Answer - Option 1 :
\(\rm \frac{2}{3}\)
Concept:
sin2 x + cos2 x = 1
\(\rm \frac{d(sinx)}{dx} = cos x\)
\(\rm \int x^ndx = \frac{x^{n + 1}}{n + 1}\)
\(\rm \int dx = x\)
Calculation:
I = \(\rm \int_{0}^{\pi/2 } cos^3 x dx\)
= \(\rm \int_{0}^{\pi/2 } cos^2 x .cos x dx\)
= \(\rm \int_{0}^{\pi/2 } (1 - sin^2 x) .cos x dx\)
Let, sin x = u
cos x dx = du
= \(\rm \int_{0}^{1} (1 - u^2)du\)
= \(\rm \left [u - \frac{u^{3}}{3} \right ]_{0}^{1}\)
= 1 - 0 - \(\rm \frac{1}{3} + 0\)
= \(\rm \frac{2}{3}\)