Question

If ${\int }_0^1\frac{\sin t}{1+t}dt=\alpha$. then value of
$I={\int }_{4\pi -2}^{4\pi }\frac{\sin \left(x/2\right)}{4\pi +2-x}dx$

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

Answer 1

The correct option is B $-\alpha$

Substitute $\frac{x}{2}=\theta$

$I={\int }_{2\pi -1}^{2\pi }\frac{\sin \theta }{4\pi +2-2\theta }2d\theta ={\int }_{2\pi -1}^{2\pi }\frac{\sin \theta }{2\pi +1-\theta }d\theta$

Now, substitute $2\pi -\theta =t,$ so that

$I={\int }_1^0\frac{\sin (2\pi -t)}{1+t}(-1)dt$

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