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\(\int\limits_0^\pi {x({{\sin }^2}(\sin x) + {{\cos }^2}(\cos x))\,dx = } \)
1. π2
2. \(\frac{{{\pi ^2}}}{2}\)
3. \(\frac{{{\pi ^2}}}{4}\)
4. \(\frac{{{\pi ^2}}}{8}\)

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Correct Answer - Option 2 : \(\frac{{{\pi ^2}}}{2}\)

I = \(\int\limits_0^π {x({{\sin }^2}(\sin x) + {{\cos }^2}(\cos x))\,dx \Rightarrow } \:I = \int\limits_0^π {(π - x)({{\sin }^2}(\sin x) + {{\cos }^2}(\cos x))\,dx} \)

Adding \(2{\rm I} = π \,\int\limits_0^π {\left( {{{\sin }^2}(\sin x)\, + \,{{\cos }^2}(\cos x)} \right)dx} \)

\(2{\rm I} = 2π \int\limits_0^{π /2} {({{\sin }^2}(\sin x) + {{\cos }^2}(\cos x))\,dx} \)

I = \(π \int\limits_0^{π /2} {({{\sin }^2}(\sin x) + {{\cos }^2}(\cos x))\,dx} \)

Also I = π \(\int\limits_0^{π /2} {[{{\sin }^2}(\cos x) + {{\cos }^2}(\sin x)]\,dx} \)

Adding \(2I = \pi \int\limits_0^{\pi /2} {2dx = 2\pi .\frac{\pi }{2} = {\pi ^2}} \Rightarrow I = \frac{{{\pi ^2}}}{2}\)

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