Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^{2} x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^{2}}$. The value of $\frac{16}{\pi^{3}} \int_{0}^{\frac{\pi}{2}} f(x) g(x) d x$ is ____.

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Definite Integrals

Let

f : [0, \frac{π}{2}] \to [0, 1]

be the function defined by

f (x) = sin^{2} x

and let

g : [0, \frac{π}{2}] \to [0, \infty)

be the function defined by

g (x) = \frac{π x}{2} - x^{2}

.

The value of

\frac{16}{π ^{3}} \int_{0}^{\frac{π}{2}} f (x) g (x) d x

is ____.

Answer: 0.25

From the previous result,

2 I_{1} = \int_{0}^{π /2} g (x) d x

, so:

I_{1} = \frac{1}{2} \int_{0}^{π /2} (\frac{π}{4})^{2} - (x - \frac{π}{4})^{2} d x

Using the standard integral

\int a^{2} - u^{2} d u = \frac{u a ^{2} - u ^{2}}{2} + \frac{a ^{2}}{2} sin^{- 1} (\frac{u}{a}) + C

with

u = x - \frac{π}{4}

and

a = \frac{π}{4}

I_{1} = \frac{1}{2} [\frac{( x - \frac{π}{4} ) \frac{π x}{2} - x ^{2}}{2} + \frac{π ^{2} /16}{2} sin^{- 1} (\frac{x - \frac{π}{4}}{\frac{π}{4}})]_{0}^{π /2}

Evaluating at the limits:

I_{1} = \frac{1}{2} (\frac{π ^{3}}{128} + \frac{π ^{3}}{128}) = \frac{π ^{3}}{128}

Therefore:

\frac{16}{π ^{3}} \cdot I_{1} = \frac{16}{π ^{3}} \cdot \frac{π ^{3}}{128} = \frac{16}{128} = \frac{1}{4} = 0.25

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Numerical answer