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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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