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(B) Let $I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} dx$. Substitute $\sqrt{x} = t$. Then,differentiating both sides with respect to $x$,we get $\frac{1}{

Vedclass · Accepted Answer

$2\sin \sqrt{x} + c$

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(B) Let $I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} dx$. Substitute $\sqrt{x} = t$. Then,differentiating both sides with respect to $x$,we get $\frac{1}{2\sqrt{x}} dx = dt$,which implies $\frac{1}{\sqrt{x}} dx = 2 dt$. Substituting these into the integral,we get $I = \int \cos(t) \cdot 2 dt = 2 \int \cos(t) dt$. Integrating $\cos(t)$,we get $I = 2 \sin(t) + c$. Substituting back $t = \sqrt{x}$,we get $I = 2 \sin \sqrt{x} + c$.

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

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