COMEDK · Maths · 30. Definite Integration
If \(I_{1}=\int_{0}^{\pi / 2} x \sin x d x\) and \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\), then which one of the following is true?
- A \(I_{1}=I_{2}\)
- B \(I_{1}+I_{2}=0\)
- C \(I_{1}=\frac{\pi}{2} I_{2}\)
- 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
\(I_{1}-\int_{0}^{\pi / 2} x \sin x d x\) \(=\left[-x \cos x+\int \cos x\right]_{0}^{\pi / 2}\) \(=[-x \cos x+\sin x]_{0}^{\pi / 2}\) \(=0+\sin \frac{\pi}{2}-0=1\) Similarly, \(I_{2}=\int_{0}^{\pi / 2} x \cos x d x\) \(\quad=\left[x \sin x-\int_{\sin x]_{0}^{\pi / 2}}\right.\)…
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