# Question 24fd5

Apr 20, 2017

$\frac{\pi}{4.}$

#### Explanation:

Respected Eddie has suggested the Method to solve the

Problem.

Here is another way to Solve it.

For that, we need the following useful Result :

Result :int_0^a f(x)dx=int_0^af(a-x)dx; a>0, and, f" is cont. on "[0,a].#

So, $I = {\int}_{0}^{\frac{\pi}{2}} {\sin}^{2} x \mathrm{dx} \ldots \ldots \left(1\right)$

$\Rightarrow I = {\int}_{0}^{\frac{\pi}{2}} {\sin}^{2} \left(\frac{\pi}{2} - x\right) \mathrm{dx} = {\int}_{0}^{\frac{\pi}{2}} {\cos}^{2} x \mathrm{dx} \ldots \ldots \left(2\right)$

$\therefore , \left(1\right) + \left(2\right) \Rightarrow I + I = 2 I = {\int}_{0}^{\frac{\pi}{2}} \left({\sin}^{2} x + {\cos}^{2} x\right) \mathrm{dx} , i . e . ,$

$2 I = {\int}_{0}^{\frac{\pi}{2}} \left(1\right) \mathrm{dx} = {\left[x\right]}_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$

$\therefore I = \frac{\pi}{4.}$

Enjoy Maths.!