Answer
$f(t)=\displaystyle \frac{8}{105}t^{7/2}+2\sin t+C_{1}t^{2}+Dt+E$
Work Step by Step
$\left[\begin{array}{ll}
\text{}f''' & \text{particular antiderivative,}f''\\
\sqrt{t}=t^{1/2} & \frac{t^{1/2+1}}{1/2+1}=\frac{2}{3}t^{3/2}\\
-2\cos t & -2\sin t
\end{array}\right]$
$f''(t)=\displaystyle \frac{2}{3}t^{3/2}+2\sin t +C$
$\left[\begin{array}{ll}
\text{}f'' & \text{particular antiderivative,}f'\\
\frac{2}{3}t^{3/2} & \frac{2}{3}\cdot\frac{t^{3/2+1}}{3/2+1}=\frac{2}{3}\cdot\frac{2}{5}t^{5/2}\\
- 2\sin t & +2\cos t\\
+C & Ct
\end{array}\right]$
$f'(t)=\displaystyle \frac{4}{15}t^{5/2}-2\cos t+Ct+D$
$\left[\begin{array}{ll}
\text{}f' & \text{particular antiderivative,}f\\
\frac{4}{15}t^{5/2} & \frac{4}{15}\cdot\frac{t^{5/2+1}}{5/2+1}=\frac{4}{15}\cdot\frac{2}{7}t^{7/2}\\
+2\cos t & +2\sin t\\
+Ct & C\cdot\frac{t^{1+1}}{1+1}=\frac{Ct^{2}}{2}=C_{1}t^{2}\\
+D & Dt
\end{array}\right]$
$f(t)=\displaystyle \frac{8}{105}t^{7/2}+2\sin t+C_{1}t^{2}+Dt+E$
