Question

If ${\int }_0^1\frac{\sin t}{1+t}dt=\alpha$, then the value of the integral ${\int }_{4\pi -2}^{4\pi }\frac{\sin t/2}{4\pi +2-t}dt$ in terms of $\alpha$ is given by

• A. $2\alpha$
• B. $-2\alpha$
• C. $\alpha$
• D. $-\alpha$

Answer 1

Answer: (D)

$-\alpha$

${\int }_{4\pi -2}^{4\pi }\frac{\sin t/2}{4\pi +2-t}dt=\frac{1}{2}{\int }_{4\pi -2}^{4\pi }\frac{\sin t/2}{1+(2\pi -\frac{t}{2})}dt$

$=2.\frac{1}{2}{\int }_0^1\frac{\sin (2\pi -u)}{1+u}du$

$($ Substitute $2\pi -\frac{t}{2}=u\Rightarrow dt=-2du)$

$={\int }_0^1\frac{\sin u}{1+u}du=-{\int }_0^1\frac{\sin t}{1+t}dt=-\alpha$