Question
$\int\limits^{\frac{\pi}{2}}_0\text{x}\sin\text{x dx}$ is equal to:
- $\frac{\pi}{4}$
- $\frac{\pi}{2}$
- $\pi$
- $1$
Solution:
We have,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{x}\sin\text{x dx}$
$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_01(-\cos\text{x})\text{dx}$
$=\big[-\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\int\limits^{\frac{\pi}{2}}_0\cos\text{x dx}$
$=-\big[\text{x}\cos\text{x}\big]^{\frac{\pi}{2}}_0+\big[\sin\text{x}\big]^{\frac{\pi}{2}}_0$
$=-\big[0-0\big]+\big[1-0\big]$
$=1$
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