The value of \(\int \sin \sqrt{x} d x\) is equal to

\(\begin{array}{ll} & I=\int \sin \sqrt{x} \mathrm{~d} x \\ \quad & \text { Let } \sqrt{x}=\mathrm{t} \\ \therefore \quad & \frac{1}{2 \sqrt{x}} \mathrm{~d} x=\mathrm{dt} \\ \therefore \quad & \mathrm{d} x=2 \sqrt{x} \mathrm{dt}=2 \mathrm{tdt} \\ \therefore \quad & \mathrm{I}=\int \sin \mathrm{t} \cdot 2 \mathrm{t} \cdot \mathrm{dt} \\ = & 2 \int \sin \mathrm{t} \cdot \mathrm{tdt}\end{array}\)
\(\begin{aligned} & =2\left[\mathrm{t} \int \sin \mathrm{t} d t-\int\left(\frac{\mathrm{d}}{\mathrm{dt}} \mathrm{t} \int \sin \mathrm{t}\right) \mathrm{dt}\right] \\ & =2\left[\mathrm{t}(-\cos t)-\int(-\cos \mathrm{t}) \mathrm{dt}\right] \\ & =2\left[-\mathrm{t} \cos t+\int \cos \mathrm{tdt}\right] \\ & =-2 \mathrm{t} \cos t+2 \sin t+c \\ & =2 \sin \sqrt{x}-2 \sqrt{x} \cos \sqrt{x}+c\end{aligned}\)