KCET · Maths · Definite Integration
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?
- A \(I_{1}=I_{2}\)
- B \(I_{1}+I_{2}=0\)
- C \(I_{1}=\frac{\pi}{2} \cdot I_{2}\)
- D \(I_{1}+I_{2}=\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(D) \(I_{1}+I_{2}=\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
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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