Question

If$\underset{0}{\overset{1}{\int }}\frac{\sin t}{1+t}dt=\alpha$, then the value of the integral $\underset{4\pi }{\overset{4\pi -2}{\int }}\frac{\sin \frac{t}{2}}{4\pi +2-t}dt$ is

• A. $\frac{\alpha }{2}$
• B. $4\alpha$
• C. $2\alpha$
• D. $\alpha$

Answer 1

The correct option is D $\alpha$

$I=\underset{4\pi }{\overset{4\pi -2}{\int }}\frac{\sin \frac{t}{2}}{4\pi +2-t}dt\Rightarrow I=\frac{1}{2}\underset{4\pi }{\overset{4\pi -2}{\int }}\frac{\sin \frac{t}{2}}{1+(2\pi -\frac{t}{2})}dt$
Put $2\pi -\frac{t}{2}=z$
$\Rightarrow \frac{1}{2}dt=-dz$
When $t=4\pi -2,z=2\pi -2\pi +1=1$
When $t=4\pi ,z=2\pi -2\pi =0$
$\Rightarrow I=-\underset{0}{\overset{1}{\int }}\frac{\sin (2\pi -z)\left(dz\right)}{1+z}\Rightarrow I=\underset{0}{\overset{1}{\int }}\frac{\sin zdz}{z+1}=\underset{0}{\overset{1}{\int }}\frac{\sin t}{1+t}dt\therefore I=\alpha$