The value of int sin sqrt{x} , dx is equal to | vedclass

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(A) Let $I = \int \sin \sqrt{x} \, dx$.Substitute $\sqrt{x} = t$,then $x = t^2$,which implies $dx = 2t \, dt$.Substituting these into the integral:$I

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$2 \sin \sqrt{x} - 2 \sqrt{x} \cos \sqrt{x} + c$,where $c$ is a constant of integration.

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(A) Let $I = \int \sin \sqrt{x} \, dx$.Substitute $\sqrt{x} = t$,then $x = t^2$,which implies $dx = 2t \, dt$.Substituting these into the integral:$I = \int \sin t \cdot (2t) \, dt = 2 \int t \sin t \, dt$.Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = \sin t \, dt$:$I = 2 \left[ t(-\cos t) - \int 1 \cdot (-\cos t) \, dt \right]$$I = 2 [ -t \cos t + \int \cos t \, dt ]$$I = 2 [ -t \cos t + \sin t ] + c$$I = 2 \sin t - 2t \cos t + c$.Substituting back $t = \sqrt{x}$:$I = 2 \sin \sqrt{x} - 2 \sqrt{x} \cos \sqrt{x} + c$.

The value of $\int \sin \sqrt{x} \, dx$ is equal to

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