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Question: Given \(\int_{0}^{1}\frac{\sin t}{1 + t}\)dt = a, find the value of\(\int_{4\pi - 2}^{4\pi}\frac{\si...

Given ∫01sin⁡t1+t\int_{0}^{1}\frac{\sin t}{1 + t}dt = a, find the value of∫4π−24πsin⁡(t/2)4π+2−t\int_{4\pi - 2}^{4\pi}\frac{\sin(t/2)}{4\pi + 2 - t} in terms of a -

A

a

B

– a

C

2a

D

None of these

Answer

– a

Explanation

Solution

I = ∫4π−24πsin⁡(t/2)4π+2−t\int_{4\pi - 2}^{4\pi}\frac{\sin(t/2)}{4\pi + 2 - t}dt

= (4p – (4p – 2)) ∫01sin⁡((4π−(4π−2))t+4π−22)4π+2−((4π−(4π−2)t+4π−2)\int_{0}^{1}\frac{\sin\left( \frac{(4\pi - (4\pi - 2))t + 4\pi - 2}{2} \right)}{4\pi + 2 - ((4\pi - (4\pi - 2)t + 4\pi - 2)}dt

= 2 ∫01sin⁡(t−1)(4−2t)dt\int_{0}^{1}{\frac{\sin(t - 1)}{(4 - 2t)}dt}

= ∫01sin⁡(t−1)(2−t)\int_{0}^{1}\frac{\sin(t - 1)}{(2 - t)}dt

= ∫01sin⁡(1−t−1)2−(1−t)\int_{0}^{1}\frac{\sin(1 - t - 1)}{2 - (1 - t)} (By Prop. IV)

= ∫01sin⁡(−t)(1+t)\int_{0}^{1}\frac{\sin( - t)}{(1 + t)}dt

= – ∫01sin⁡t1+t\int_{0}^{1}\frac{\sin t}{1 + t}dt

= – a. (given).