Paragraph: Let \(f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]\) be the function defined by \(f(x)=\sin…
Paragraph: Let \(f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]\) be the function defined by \(f(x)=\sin ^2 x\) and
let $g: [0, \frac{\pi}{2}] \rightarrow [0, \infty)$be the function defined by \(g(x)=\sqrt{\frac{\pi x}{2}-x^2}\).
Question: The value of $2 \int_0^{\frac{\pi}{2}} f(x) g(x) d x-\int_0^{\frac{\pi}{2}} g(x) d x$ is
Solution
$I=2 \int_0^{\frac{\pi}{2}} \underbrace{\sin ^2 x \cdot \sqrt{\frac{\pi x}{2}-x^2}}_{I_1}-\int_0^{\frac{\pi}{2}} g(x) d x$
Let $I_1=\int_0^{\frac{\pi}{2}} \sin ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2}$
(making perfect square)
apply kings
$I_1=\int_0^{\frac{\pi}{2}} \cos ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(\frac{\pi}{2}-x\right)^2}$
add both
$2 I_1=\int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2}$
i.e. $2 I_1=\int_0^{\frac{\pi}{2}} g(x)$
Now $I=2 I_1-\int_0^{\frac{\pi}{2}} g(x)=0$