Set $\displaystyle x = 7\sin(u)$.
Then $\displaystyle dx = 7\cos(u)du$.
So our integral becomes...
$\displaystyle \int_0^{\arcsin(\tfrac{1}{7}p)} \sqrt{49 - 49\sin^2(u)} 7\cos(u) du$
$\displaystyle = \int_0^{\arcsin(\tfrac{1}{7}p)} 49\sqrt{\cos^2(u)} \cos(u) du$
$\displaystyle = \int_0^{\arcsin(\tfrac{1}{7}p)} 49\cos^2(u) du$
$\displaystyle = \bigg{[} 49 \bigg{(} \frac{u}{2} + \frac{1}{4}\sin(2u) \bigg{)}\bigg{]}_0^{\arcsin(\tfrac{1}{7}p)}$
$\displaystyle = 49 \bigg{(} \frac{\arcsin(\tfrac{1}{7}p)}{2} + \frac{1}{4}\sin(2\arcsin(\tfrac{1}{7}p)) \bigg{)}$
Posting working as I go along...
Evaluate the integral and solve for P
let $\displaystyle x = 7sin \theta $
So our bounds become
upper: $\displaystyle \arcsin \frac{p}{7}$
lower: $\displaystyle 0$
$\displaystyle dx = 7cos \theta d \theta$
We will leave the limits for now.
$\displaystyle \int \sqrt{7^2 - 7^2 sin^2 \theta} (7cos \theta) d \theta $
$\displaystyle 49 \int \sqrt{1 - 1 sin^2 \theta} (cos \theta) d \theta $
$\displaystyle 49 \int \sqrt{cos^2 \theta } (cos \theta) d \theta $
$\displaystyle 49 \int cos^2 d \theta $
$\displaystyle 49 \int \frac{1-cos2x}{2} d \theta $
$\displaystyle \frac{49}{2} \int d \theta - \frac{49}{2} \int cos2x d \theta $
$\displaystyle \frac{49}{2} \theta + \frac{49}{4} sin2 \theta$
Remember
$\displaystyle \theta=\arcsin \frac{x}{7}$
$\displaystyle \frac{49}{2} \arcsin \frac{x}{7}+ \frac{49}{4} sin(2 \arcsin \frac{x}{7}) $
This is evaluated from 0--P so we have
$\displaystyle \frac{49}{2} \arcsin \frac{p}{7}+ \frac{49}{4} sin(2 \arcsin \frac{p}{7}) $
Let us note that any combination of $\displaystyle n \pi $ will make sin = 0 but we will take n=0 for simplicity.
$\displaystyle \frac{49}{2} \arcsin \frac{p}{7}+ \frac{49}{4} sin(2 \arcsin \frac{p}{7}) = \frac{49 \pi}{12} $
Solve for P
Ugg ok scrap that use this instead...
$\displaystyle \int \cos^2(x) dx = \frac{1}{2}\cos(x)\sin(x)+\frac{1}{2}x$
So we get instead...
$\displaystyle = \int_0^{\arcsin(\tfrac{1}{7}p)} 49\cos^2(u) du = \bigg{[} \frac{49}{2}\cos(u)\sin(u)+\frac{49}{2}u \bigg{]}_0^{\arcsin(\tfrac{1}{7}p)}$
$\displaystyle = \frac{49}{2}\cos(\arcsin(\tfrac{1}{7}p))\sin(\arcs in(\tfrac{1}{7}p))+\frac{49}{2}\arcsin(\tfrac{1}{7 }p)$
$\displaystyle = \frac{49}{2} \cdot \frac{1}{7} \cdot \sqrt{49-p^2} \cdot \frac{p}{7}+\frac{49}{2}\arcsin(\tfrac{1}{7}p)$
$\displaystyle = \frac{p}{2} \cdot \sqrt{49-p^2} + \frac{49}{2}\arcsin(\tfrac{1}{7}p)$
Then...
Set this equal to...
$\displaystyle \frac{49\pi}{12}$...
To get this hellish thing to solve.
$\displaystyle = \frac{p}{2} \cdot \sqrt{49-p^2} + \frac{49}{2}\arcsin(\tfrac{1}{7}p) = \frac{49\pi}{12}$
=> $\displaystyle = 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) = 49\pi$
...
Using Maple now since I don't know how to simplify this off the top of my head you get
$\displaystyle
p=1.854524594$
See attached images.
First one is the graph of
$\displaystyle 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) - 49\pi$
We want to know where this one crosses the x-axis.
Second one the graphs of of...
$\displaystyle 6p \cdot \sqrt{49-p^2} + 294\arcsin(\tfrac{1}{7}p) \textrm{ and } 49\pi$
With the blue line representing $\displaystyle 49\pi$.
We want to see where they cross.
This value is the p value I posted above. Maybe Opalg or Krizalid could give you a better answer but this is the best I can do for now.
As the curve is a semicircle, the integral is the sum of sector area and a triangle area.
$\displaystyle sin\left(\frac{{\pi}}{2}-\theta\right)=\frac{P}{7}$
$\displaystyle \frac{{\pi}}{2}-\theta=sin^{-1}\left(\frac{P}{7}\right)$
The integral is
$\displaystyle \frac{49}{2}sin^{-1}\left(\frac{P}{7}\right)+\frac{P}{2}\sqrt{49-P^2}$
then use the intersection of curves and lines or other methods for P.