Question

By using the properties of definite integrals, evaluate the integral ${\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$

Answer 1

$I={\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$ .............. (1)
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}{\cos }^2(\frac{\pi }{2}-x)dx,(\because {\int }_0^{0}f(x)dx={\int }_0^{0}f(a-x\left)dx\right)$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx$ .......... (2)
Adding (1) and (2), we obtain
$2I={\int }_0^{\frac{\pi }{2}}({\sin }^{2}x+{\cos }^{2}x)dx$
$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}1\cdot dx$
$\Rightarrow 2I=[x{]}_0^{\frac{\pi }{2}}$
$\Rightarrow 2I=\frac{\pi }{2}$
$\Rightarrow I=\frac{\pi }{4}$