int sin sqrt{x} ,d x=ldots+C (where C is a co | vedclass

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(A) To evaluate $\int \sin \sqrt{x} \,d x$,let $\sqrt{x} = t$. Then $x = t^2$,which implies $d x = 2t \,d t$. Substituting these into the integral,we

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$2(-\sqrt{x} \cos \sqrt{x}+\sin \sqrt{x})$

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(A) To evaluate $\int \sin \sqrt{x} \,d x$,let $\sqrt{x} = t$. Then $x = t^2$,which implies $d x = 2t \,d t$. Substituting these into the integral,we get: $\int \sin \sqrt{x} \,d x = \int \sin t (2t \,d t) = 2 \int t \sin t \,d t$. Using integration by parts,where $\int u \,dv = uv - \int v \,du$,let $u = t$ and $dv = \sin t \,d t$. Then $du = d t$ and $v = -\cos t$. $2 \int t \sin t \,d t = 2 \left[ t(-\cos t) - \int (-\cos t) \,d t \right]$ $= 2 [-t \cos t + \int \cos t \,d t]$ $= 2 [-t \cos t + \sin t] + C$. Substituting $t = \sqrt{x}$ back,we get: $= 2(-\sqrt{x} \cos \sqrt{x} + \sin \sqrt{x}) + C$.

$\int \sin \sqrt{x} \,d x=\ldots+C$ (where $C$ is a constant of integration.)

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