Let f:[0, \frac\pi2] arrow[0,1] be the function defined by f(x)=\sin ^2 x and let g:[0, \frac\pi2] arrow[0, \infty) be the function defined by g(x)=\sqrt\frac\pi x2-x^2 . The value of \frac16\pi^3 \int\limits_0^\frac\pi2 f(x) g(x) d x is \underline\hspace2cm .

Definite Integration PYQ — JEE Advanced 2024 · Shift 2 · Q34 with Solution

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Correct answer: 0.25

We need to evaluate $I=\frac{16}{\pi^3}\int_0^{\pi/2} \sin^2 x\,\sqrt{\frac{\pi x}{2}-x^2}\,dx.$
First simplify the square root: $\frac{\pi x}{2}-x^2=x\left(\frac\pi2-x\right).$ So the integral is $I=\frac{16}{\pi^3}\int_0^{\pi/2} \sin^2 x\,\sqrt{x\left(\frac\pi2-x\right)}\,dx.$
Use the identity $\sin^2 x=\frac{1-\cos 2x}{2}.$ Hence $I=\frac{16}{\pi^3}\cdot \frac12\int_0^{\pi/2} (1-\cos 2x)\sqrt{x\left(\frac\pi2-x\right)}\,dx.$ That is, $I=\frac{8}{\pi^3}\left[\int_0^{\pi/2} \sqrt{x\left(\frac\pi2-x\right)}\,dx-\int_0^{\pi/2} \cos 2x\,\sqrt{x\left(\frac\pi2-x\right)}\,dx\right].$
Now use symmetry about $x=\frac\pi4$ . Let $J=\int_0^{\pi/2} \cos 2x\,\sqrt{x\left(\frac\pi2-x\right)}\,dx.$ Substitute $x\mapsto \frac\pi2-x$ . Then $\cos 2\left(\frac\pi2-x\right)=\cos(\pi-2x)=-\cos 2x,$ and $\sqrt{\left(\frac\pi2-x\right)x}=\sqrt{x\left(\frac\pi2-x\right)}.$ Therefore the integrand changes sign, so $J=-J \implies J=0.$ Thus $I=\frac{8}{\pi^3}\int_0^{\pi/2} \sqrt{x\left(\frac\pi2-x\right)}\,dx.$
Let $a=\frac\pi2.$ Then we need $\int_0^a \sqrt{x(a-x)}\,dx.$ A standard result is $\int_0^a \sqrt{x(a-x)}\,dx=\frac{\pi a^2}{8}.$ Using $a=\frac\pi2$ ,

=\frac{\pi}{8}\left(\frac\pi2\right)^2 =\frac{\pi}{8}\cdot \frac{\pi^2}{4} =\frac{\pi^3}{32}.$$ 6. Therefore, $$I=\frac{8}{\pi^3}\cdot \frac{\pi^3}{32}=\frac14.$$ 7. Hence the required value is $$\boxed{\frac14}.$$ Since this is an integer-type entry format but the stored answer is decimal, the value is $0.25$.

View written solutionFree

Correct answer: 0.25

We need to evaluate $I=\frac{16}{\pi^3}\int_0^{\pi/2} \sin^2 x\,\sqrt{\frac{\pi x}{2}-x^2}\,dx.$
First simplify the square root: $\frac{\pi x}{2}-x^2=x\left(\frac\pi2-x\right).$ So the integral is $I=\frac{16}{\pi^3}\int_0^{\pi/2} \sin^2 x\,\sqrt{x\left(\frac\pi2-x\right)}\,dx.$
Use the identity $\sin^2 x=\frac{1-\cos 2x}{2}.$ Hence $I=\frac{16}{\pi^3}\cdot \frac12\int_0^{\pi/2} (1-\cos 2x)\sqrt{x\left(\frac\pi2-x\right)}\,dx.$ That is, $I=\frac{8}{\pi^3}\left[\int_0^{\pi/2} \sqrt{x\left(\frac\pi2-x\right)}\,dx-\int_0^{\pi/2} \cos 2x\,\sqrt{x\left(\frac\pi2-x\right)}\,dx\right].$
Now use symmetry about $x=\frac\pi4$ . Let $J=\int_0^{\pi/2} \cos 2x\,\sqrt{x\left(\frac\pi2-x\right)}\,dx.$ Substitute $x\mapsto \frac\pi2-x$ . Then $\cos 2\left(\frac\pi2-x\right)=\cos(\pi-2x)=-\cos 2x,$ and $\sqrt{\left(\frac\pi2-x\right)x}=\sqrt{x\left(\frac\pi2-x\right)}.$ Therefore the integrand changes sign, so $J=-J \implies J=0.$ Thus $I=\frac{8}{\pi^3}\int_0^{\pi/2} \sqrt{x\left(\frac\pi2-x\right)}\,dx.$
Let $a=\frac\pi2.$ Then we need $\int_0^a \sqrt{x(a-x)}\,dx.$ A standard result is $\int_0^a \sqrt{x(a-x)}\,dx=\frac{\pi a^2}{8}.$ Using $a=\frac\pi2$ ,

=\frac{\pi}{8}\left(\frac\pi2\right)^2 =\frac{\pi}{8}\cdot \frac{\pi^2}{4} =\frac{\pi^3}{32}.$$ 6. Therefore, $$I=\frac{8}{\pi^3}\cdot \frac{\pi^3}{32}=\frac14.$$ 7. Hence the required value is $$\boxed{\frac14}.$$ Since this is an integer-type entry format but the stored answer is decimal, the value is $0.25$.