Question

$If{\int }_0^1\frac{\sin t}{1+t}dt=\alpha ,ifvalueofl={\int }_{4\pi -2}^{4\pi }\frac{\sin \left(x/2\right)}{4\pi +2-x}dxis\frac{-k}{4}\alpha ,thenkisequalto.$

Answer 1

$I=\underset{4\pi -2}{\overset{4\pi }{\int }}\frac{\sin t/2}{2(1+(2\pi -t/2\left)\right)}dt$

$2\pi -t/2=x$

$-\frac{dt}{2}=dx$

$dt=-2dx$

$t\to 4\pi -2\Rightarrow x=1$

$t=4\pi \Rightarrow x=0$

$I=\frac{1}{2}{\int }_0^1\frac{\sin (2\pi -x)}{1+x}(-2dx)$

$={\int }_0^1\frac{-\sin x}{1+x}dx$

$-{\int }_0^1\frac{\sin t}{1+t}dt$

$I=-\alpha$

$\therefore =-\frac{k}{4}\alpha$

$=k=4$