Question$\int_{0}^{4} \frac{p}{\sqrt{9+p^{2}}} d p$

Studdy Solution
Back-substitute and evaluate the definite integral:
At $\theta = \tan^{-1}\left(\frac{4}{3}\right)$, $\cos(\theta) = \frac{3}{5}$ (since $\sin(\theta) = \frac{4}{5}$ and $\cos^2(\theta) + \sin^2(\theta) = 1$).
At $\theta = 0$, $\cos(0) = 1$.
Evaluate:
$-3 \left(\frac{3}{5}\right) - (-3 \cdot 1) = -\frac{9}{5} + 3 = \frac{15}{5} - \frac{9}{5} = \frac{6}{5}$
The value of the integral is:
$\boxed{\frac{6}{5}}$