Evaluate int sin (sqrt{k}) , dk on (0, infty) | vedclass

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(D) Let $I = \int \sin (\sqrt{k}) \, dk$ for $k \in (0, \infty)$.Substitute $k = t^2$,which implies $dk = 2t \, dt$.Substituting these into the integr

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$2[\sin (\sqrt{k})-\sqrt{k} \cos (\sqrt{k})]+c$

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(D) Let $I = \int \sin (\sqrt{k}) \, dk$ for $k \in (0, \infty)$.Substitute $k = t^2$,which implies $dk = 2t \, dt$.Substituting these into the integral,we get:$I = \int \sin(t) \cdot (2t \, dt) = 2 \int t \sin(t) \, dt$.Using integration by parts,where $\int u \, dv = uv - \int v \, du$,let $u = t$ and $dv = \sin(t) \, dt$. Then $du = dt$ and $v = -\cos(t)$.$I = 2 [t(-\cos(t)) - \int (-\cos(t)) \, dt]$$I = 2 [-t \cos(t) + \int \cos(t) \, dt]$$I = 2 [-t \cos(t) + \sin(t)] + c$Substituting $t = \sqrt{k}$ back into the expression:$I = 2[\sin(\sqrt{k}) - \sqrt{k} \cos(\sqrt{k})] + c$.

Evaluate $\int \sin (\sqrt{k}) \, dk$ on $(0, \infty)$.

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