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If $I_{1}=\int_{0}^{\pi / 2} x \cdot \sin x d x$ and $I_{2}=\int_{0}^{\pi / 2} x \cdot \cos x d x$, then which one of the following is true?
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Verified Answer
The correct answer is:
$I_{1}+I_{2}=\frac{\pi}{2}$
Given,
$\begin{aligned} I_{1} &=\int_{0}^{\pi / 2} x \cdot \sin x d x \text { and } I_{2}=\int_{0}^{\pi / 2} x \cos x d x \\ \text { Now, } l_{1} &=[x(-\cos x)]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} 1 \cdot(-\cos x) d x \\ &=[0+0]+[\sin x]_{0}^{\pi / 2}=1 \\ \text { and } l_{2} &=[x \cdot \sin x]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} 1 \cdot \sin x d x \\ &=\left[\frac{\pi}{2}-0\right]-[-\cos x]_{0}^{\pi / 2} \\ &=\frac{\pi}{2}+(0-1)=\frac{\pi}{2}-1 \\ \therefore & l_{1}+I_{2}=\frac{\pi}{2} \end{aligned}$
$\begin{aligned} I_{1} &=\int_{0}^{\pi / 2} x \cdot \sin x d x \text { and } I_{2}=\int_{0}^{\pi / 2} x \cos x d x \\ \text { Now, } l_{1} &=[x(-\cos x)]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} 1 \cdot(-\cos x) d x \\ &=[0+0]+[\sin x]_{0}^{\pi / 2}=1 \\ \text { and } l_{2} &=[x \cdot \sin x]_{0}^{\pi / 2}-\int_{0}^{\pi / 2} 1 \cdot \sin x d x \\ &=\left[\frac{\pi}{2}-0\right]-[-\cos x]_{0}^{\pi / 2} \\ &=\frac{\pi}{2}+(0-1)=\frac{\pi}{2}-1 \\ \therefore & l_{1}+I_{2}=\frac{\pi}{2} \end{aligned}$
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