(1) 0
\(f(x)=\sin ^2 x, g(x)=\sqrt{\frac{\pi}{2} x-x^2}\)
Here \(f\left(\frac{\pi}{2}-x\right)=\cos ^2 x, g\left(\frac{\pi}{2}-x\right)=g(x)\)
Let \(I_1=2 \int\limits_0^{\frac{\pi}{2}} f(x) g(x)=2 \int\limits_0^{\frac{\pi}{2}} \sin ^2 x \cdot g(x) d x \quad....(1)\)
as \(\int\limits_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x\)
\(\Rightarrow I_1=2 \int\limits_0^{\frac{\pi}{2}} \cos ^2 x g(x) d x\quad....(2)\)
\((1)+(2)\)
\(\Rightarrow 2 I_1=2 \int\limits_0^{\frac{\pi}{2}} g(x) d x\)
\(\Rightarrow I_1=\int\limits_0^{\frac{\pi}{2}} g(x) d x\)
\(\Rightarrow 2 \int\limits_0^{\frac{\pi}{2}} f(x) g(x)-\int\limits_0^{\frac{\pi}{2}} g(x) d x=0\)
(2) 0.25
\(2 \int\limits_0^{\frac{\pi}{2}} f(x) g(x) d x=\int\limits_0^{\frac{\pi}{2}} g(x) d x=I_1 \text { (let) }\)
Now, \( I_1=\int\limits_0^{\frac{\pi}{2}} g(x) d x=\int\limits_0^{\frac{\pi}{2}} \sqrt{\frac{\pi}{2} x-x^2} \ d x\)
\(I_1=\int\limits_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(\frac{\pi}{4}-x\right)^2} d x\)
Put \(\frac{\pi}{4}-x=t\)
\(\Rightarrow d x=-d t\)
\(I_1=-\int\limits_{\frac{\pi}{4}}^{\frac{-\pi}{4}} \sqrt{\left(\frac{\pi}{4}\right)^2-t^2} d t\)
\(I_1=\int\limits_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sqrt{\left(\frac{\pi}{4}\right)^2-t^2} d t\)
\(I_1=2 \int\limits_0^{\frac{\pi}{4}} \sqrt{\left(\frac{\pi}{4}\right)^2-t^2} d t=2\left[\frac{t}{2} \sqrt{\left(\frac{\pi}{4}\right)^2-t^2}+\frac{\pi^2}{32} \sin ^{-1}\left(\frac{4 t}{\pi}\right)\right]_0^{\frac{\pi}{4}}\)
\(I_1=\frac{\pi^3}{32}\)
Now,
\(I=\frac{8}{\pi^3}I_1\)
\(I=\frac{1}{4}=0.25\)
