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Answer: $\frac{\pi }{2}-1$

Hint:Use indefinite formula then put the limit to solve this integral

Given: $\int_{0}^{\frac{\pi }{2}}x\cos xdx$

Solution:

$\int_{0}^{\frac{\pi }{2}}x\cos xdx$

Integrating by parts  =>  Let x be the 1st part and cos x be the 2nd part

\begin{aligned} &=\left[x \int \cos x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d(x)}{d x} \int \cos x d x\right\} d x \\ &=[x \sin x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}(1 \cdot \sin x) d x \end{aligned} \quad\left[\int \cos x d x=\sin x\right]

$=\left[\frac{\pi}{2} \sin \frac{\pi}{2}-0 \times \sin 0\right]-\int_{0}^{\frac{\pi}{2}} \sin x d x \quad\left[\sin \frac{\pi}{2}=1, \sin 0=0\right]$

\begin{aligned} &=\left[\frac{\pi}{2} \cdot 1-0\right]-[-\cos x]_{0}^{\frac{\pi}{2}} \; \; \; \; \; \; \; \; \; \quad\left[\int \sin x d x=-\cos x\right] \\ &=\frac{\pi}{2}+[\cos x]_{0}^{\frac{\pi}{2}} \\ &=\frac{\pi}{2}+\left[\cos \frac{\pi}{2}-\cos 0\right] \end{aligned} \quad\left[\cos \frac{\pi}{2}=0, \cos 0=1\right]

$=\frac{\pi }{2}+\left [ 0-1 \right ]$

$=\frac{\pi }{2}-1$

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