Question

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

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

Answer 1

Answer: (D)

$-\alpha$

$I={\int }_{4\pi -2}^{4\pi }\frac{\sin \frac{t}{2}}{4\pi +2-t}dt$
$=\frac{1}{2}{\int }_{4\pi -2}^{4\pi }\frac{\sin \frac{t}{2}}{1+(2\pi -\frac{t}{2})}dt$
Put, $2\pi -\frac{t}{2}=z$
$\therefore -\frac{1}{2}dt=dz,i.e.,dt=-2dz$
when $t=4\pi -2,z=2\pi -2\pi +1=1$
when $t=4\pi ,z=2\pi -2\pi =0$
$\therefore I=\frac{1}{2}{\int }_1^0\frac{\sin (2\pi -z)(-2dz)}{1+z}$
$={\int }_0^1\frac{-\sin zdz}{z+1}=-\int \frac{\sin t}{1+t}dt=-\alpha$