Tangent line of $y=A(x)$ at $x=\pi / 2$, where $A(x)$ is an integral function 
Given the function $A$ defined by: 
  $$A(x) = \int^x_{\pi/x}\frac{\sin t}{t} \,dt$$
  Find the equation of the tangent line of $y=A(x)$ at
  $x=\frac{\pi}{2}$.

please try to explain in detail 
Thank you in advance
 A: We want to find a tangent line, and that involves finding the slope (derivative).  So, let's find $A'(x)$.
We can apply the Fundamental Theorem of Calculus (FTC), which gives the result:

$$\frac{d}{dx}\int_a^x f(t)dt  = f(x)\tag{1}$$

So, we need to break your $A(x)$ into a form that matches $(1)$.  Basically, we want a constant in the lower limit of integration, and a function of $x$ to be in the upper limit of integration.  So, we have:
$$\begin{align}
A(x) &= \int^x_{\pi/x}\frac{\sin t}{t} \,dt\\
 &= \int^a_{\pi/x}\frac{\sin t}{t} \,dt + \int^x_a\frac{\sin t}{t} \,dt \\
 &= -\int_a^{\pi/x}\frac{\sin t}{t} \,dt + \int^x_a\frac{\sin t}{t} \,dt \\
\end{align}$$
Now, we can apply the FTC to this to find $A'(x)$:
$$\begin{align}A'(x) &= \frac{d}{dx}\left[-\int_a^{\pi/x}\frac{\sin t}{t} \,dt\right] + \frac{d}{dx}\left[\int^x_a\frac{\sin t}{t} \,dt\right] \\
&= \frac{d}{dx}\left[-\int_a^{\pi/x}\frac{\sin t}{t} \,dt\right] + \left[\frac{\sin x}{x} \right]\\
&= \underbrace{\left[\frac{\sin \left(\pi/x\right)}{\pi/x}\cdot\frac{\pi}{x^2}\right]}_\text{don't forget chain rule!} + \left[\frac{\sin x}{x} \right]\\\\
&= \left[\frac{\sin \left(\pi/x\right)}{x}\right] + \left[\frac{\sin x}{x} \right]\\
&= \frac{\sin(\pi/x) + \sin x}{x} \tag{2}
\end{align}
$$
Now, just use $(2)$ with julien's formula from the comments above, and you're basically done.  Leave a comment if you have questions.
