int cos sqrt{x} , dx is equal to | vedclass

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(A) Let $I = \int \cos \sqrt{x} \, dx$. Substitute $\sqrt{x} = t$,so $x = t^2$. Differentiating with respect to $x$,we get $dx = 2t \, dt$. Substituti

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \cos \sqrt{x} \, dx$. Substitute $\sqrt{x} = t$,so $x = t^2$. Differentiating with respect to $x$,we get $dx = 2t \, dt$. Substituting these into the integral: $I = \int \cos(t) \cdot 2t \, dt = 2 \int t \cos(t) \, dt$. Using integration by parts,where $u = t$ and $dv = \cos(t) \, dt$: $I = 2 \left[ t \sin(t) - \int \sin(t) \, dt \right] = 2 [t \sin(t) + \cos(t)] + c$. Replacing $t$ with $\sqrt{x}$: $I = 2 \sqrt{x} \sin \sqrt{x} + 2 \cos \sqrt{x} + c$.

$\int \cos \sqrt{x} \, dx$ is equal to

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