Question

If $f\left(x\right)={\int }_{-1}^1\frac{\sin x}{1+t^2}dt$ then $f^{\prime }\left(\frac{\pi }{3}\right)$ is

• A. nonexistent
• B. $\pi /4$
• C. $\pi \sqrt{3/4}$
• D. none of these

Answer 1

The correct option is B $\pi /4$

$f\left(x\right)={\int }_{-1}^1\frac{\sin x}{1+t^2}dt=\sin x{\int }_{-1}^1\frac{1}{1+t^2}dt$
$=\sin x{\left[{\tan }^{-1}t\right]}_{-1}^1=\sin x[\frac{\pi }{4}+\frac{\pi }{4}]=\sin x\left[\frac{\pi }{2}\right]$
$\therefore f^{\prime }\left(x\right)=\frac{\pi }{2}\cos x\Rightarrow f^{\prime }\left(\frac{\pi }{3}\right)=\frac{\pi }{2}\frac{1}{2}=\frac{\pi }{4}$