Question

Solve :-

$I=4{\int }_0^{\pi /2}{\sin }^{2}xdx$

Answer 1

Using the property that:
${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I=4{\int }_0^{\pi /2}{\sin }^2(\frac{\pi }{2}-x)dx=4{\int }_0^{\pi /2}{\cos }^{2}xdx$ (1)
Adding the original question to (1):
$2I=4{\int }_0^{\pi /2}({\sin }^{2}x+{\cos }^{2}x)dx=4{\int }_0^{\pi /2}1dx=4\cdot \frac{\pi }{2}$
$⟹I=\pi$