Question
Evaluate the following integrals:
$\int^\limits\frac{\pi}{2}_0\text{x}^2\sin\text{x dx}$

Answer

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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