int frac{1}{sqrt{x}} sin sqrt{x} , dx = | vedclass

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(B) To solve the integral $\int \frac{1}{\sqrt{x}} \sin \sqrt{x} \, dx$,we use the method of substitution. Let $t = \sqrt{x}$. Then,differentiating bo

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) To solve the integral $\int \frac{1}{\sqrt{x}} \sin \sqrt{x} \, dx$,we use the method of substitution. Let $t = \sqrt{x}$. Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{2\sqrt{x}}$,which implies $\frac{dx}{\sqrt{x}} = 2 \, dt$. Substituting these into the integral,we get: $\int \sin(t) \cdot 2 \, dt = 2 \int \sin(t) \, dt$. The integral of $\sin(t)$ is $-\cos(t)$. Therefore,$2(-\cos(t)) + c = -2 \cos(t) + c$. Substituting back $t = \sqrt{x}$,we get $-2 \cos \sqrt{x} + c$.

$\int \frac{1}{\sqrt{x}} \sin \sqrt{x} \, dx = $

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