Question
Evaluate: $\int_0^{\frac{\pi}{2}} x \sin x d x$
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Answer
$\text { Let } I =\int_0^{\frac{\pi}{2}} x \sin x d x $
$=\left[x \int \sin x d x\right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\left[\frac{ d }{ d x}(x) \int \sin x d x\right] d x \\ =[x(-\cos x)]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}(1)(-\cos x) d x$
$=[-x \cos x]_0^{\frac{\pi}{2}}+[\sin x]_0^{\frac{\pi}{2}} $
$=-\left(\frac{\pi}{2} \cos \frac{\pi}{2}-0\right)+\left(\sin \frac{\pi}{2}-\sin 0\right)$
$=-\left(\frac{\pi}{2} \times 0\right)+(1-0) $
$ \therefore I =1$
$=\left[x \int \sin x d x\right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\left[\frac{ d }{ d x}(x) \int \sin x d x\right] d x \\ =[x(-\cos x)]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}(1)(-\cos x) d x$
$=[-x \cos x]_0^{\frac{\pi}{2}}+[\sin x]_0^{\frac{\pi}{2}} $
$=-\left(\frac{\pi}{2} \cos \frac{\pi}{2}-0\right)+\left(\sin \frac{\pi}{2}-\sin 0\right)$
$=-\left(\frac{\pi}{2} \times 0\right)+(1-0) $
$ \therefore I =1$
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