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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

$\int \sin \sqrt{x} \, dx = $

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