Question
Evaluate the following definite integrals:
$\int\limits_{0}^\limits{\frac{\pi}{2}}\text{x}^2\sin\text{x}\text{ dx}$
$\int\limits_{0}^\limits{\frac{\pi}{2}}\text{x}^2\sin\text{x}\text{ dx}$
✓
Answer
Let $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\sin\text{x}\text{ dx}$
Apply integration by part.
$\text{I}=\big[\text{x}^2(-\cos\text{x})\big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}2\text{x}(-\cos\text{x})\text{ dx}$
$\Rightarrow\text{I}=(0-0)+2\int_{0}^\limits{\frac{\pi}{2}}\text{x}\cos\text{x}\text{ dx}$ $\Big(\cos\frac{\pi}{2}=0\Big)$
Apply integration by part again,
$\text{I}=0+2\Bigg[\big[\text{x}\sin\text{x}\big]_0^{\frac{\pi}{2}}-\int_{0}^\limits{\frac{\pi}{2}}1\times\sin\text{x}\text{ dx}\Bigg]$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}\sin\frac{\pi}{2}-0\Big)-2\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x dx}$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}-0\Big)-\big[2(-\cos\text{x})\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=\pi+2\Big(\cos\frac{\pi}{2}-\cos0\Big)$
$\Rightarrow\text{I}=\pi+2(0-1)$
$\Rightarrow\text{I}=\pi-2$
Apply integration by part.
$\text{I}=\big[\text{x}^2(-\cos\text{x})\big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}2\text{x}(-\cos\text{x})\text{ dx}$
$\Rightarrow\text{I}=(0-0)+2\int_{0}^\limits{\frac{\pi}{2}}\text{x}\cos\text{x}\text{ dx}$ $\Big(\cos\frac{\pi}{2}=0\Big)$
Apply integration by part again,
$\text{I}=0+2\Bigg[\big[\text{x}\sin\text{x}\big]_0^{\frac{\pi}{2}}-\int_{0}^\limits{\frac{\pi}{2}}1\times\sin\text{x}\text{ dx}\Bigg]$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}\sin\frac{\pi}{2}-0\Big)-2\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x dx}$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}-0\Big)-\big[2(-\cos\text{x})\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=\pi+2\Big(\cos\frac{\pi}{2}-\cos0\Big)$
$\Rightarrow\text{I}=\pi+2(0-1)$
$\Rightarrow\text{I}=\pi-2$
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