Question
Evaluate the following integrals:$\int\text{x}^2\cos\text{x dx}$
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Answer
Let $\text{I}=\int\text{x}^2\cos\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}^2\int\cos\text{x dx}-\int(2\text{x}\int\cos\text{x dx})\text{dx}$
$=\text{x}^2\sin\text{x}-2\int\text{x}\sin\text{x dx}$
$=\text{x}^2\sin\text{x}-2[\text{x}\int\sin\text{x dx}-\int(1\int\sin\text{x dx})\text{dx}]$
$=\text{x}^2\sin\text{x}-2[\text{x}(-\cos\text{x})-\int(-\cos\text{x})\text{dx}]$
$=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\int(\cos\text{x})\text{dx}$
$\text{I}=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x}+\text{C}$
Using integration by parts,
$\text{I}=\text{x}^2\int\cos\text{x dx}-\int(2\text{x}\int\cos\text{x dx})\text{dx}$
$=\text{x}^2\sin\text{x}-2\int\text{x}\sin\text{x dx}$
$=\text{x}^2\sin\text{x}-2[\text{x}\int\sin\text{x dx}-\int(1\int\sin\text{x dx})\text{dx}]$
$=\text{x}^2\sin\text{x}-2[\text{x}(-\cos\text{x})-\int(-\cos\text{x})\text{dx}]$
$=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\int(\cos\text{x})\text{dx}$
$\text{I}=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x}+\text{C}$
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