We have, Using by parts, we get $\text{x}^2\int\sin\text{x dx}-\int\big(\sin\text{x dx}\big)\frac{\text{dx}^2}{\text{dx}}\text{ dx}$ $=\text{x}^2\cos\text{x}+\int\cos\text{x }2\text{x dx}$ Again applying by parts $=\text{x}^2\cos\text{x}+2\Big[\text{x}\int\cos\text{x dx}-\int\big(\int\cos\text{x dx}\big)\frac{\text{dx}}{\text{dx}}\text{ dx}\Big]$ $=\text{x}^2\cos\text{x}+2\big[\text{x}\sin\text{x}-\int\sin\text{x dx}\big]$ $=\Big[\text{x}^2\cos\text{x}+2\text{x }\sin\text{x}+2\cos\text{x}\Big]^{\frac{\pi}{2}}_0$ $=\pi+0-0-0-2$ $=\pi-2$
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