Mathematics

If $$\int\limits_0^1 {\frac{{\sin t}}{{1 + t}}} dt = \alpha ,\,then\,\int\limits_{4\pi  - 2}^{4\pi } {\frac{{\sin (t/2)}}{{4\pi  + 2 - t}}dt} $$ in terms of $$  \alpha $$ is given by


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$$ - \alpha $$


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Single Correct Medium Published on 17th 09, 2020
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