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(A) To evaluate the integral $I = \int \cos \sqrt{x} \, dx$,we use the substitution method. Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t

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$2[\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + C$

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(A) To evaluate the integral $I = \int \cos \sqrt{x} \, dx$,we use the substitution method. Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$. Substituting these into the integral,we get: $I = \int \cos(t) \cdot (2t \, dt) = 2 \int t \cos t \, dt$. Now,we apply integration by parts,where $\int u \, dv = uv - \int v \, du$. Let $u = t$ and $dv = \cos t \, dt$. Then $du = dt$ and $v = \sin t$. $I = 2 [t \sin t - \int \sin t \, dt]$. $I = 2 [t \sin t - (-\cos t)] + C$. $I = 2 [t \sin t + \cos t] + C$. Substituting $t = \sqrt{x}$ back into the expression: $I = 2[\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + C$.

$\int \cos \sqrt{x} \, dx =$ (where $C$ is a constant of integration.)

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