∫ sin(√x) dx = ?

To solve the integral $\int \sin(\sqrt{x})\,dx$, we will use a substitution followed by integration by parts.

Step 1: Perform a substitution. Let $u = \sqrt{x}$. Then $u^2 = x$. Differentiating both sides with respect to $u$ gives $2u\,du = dx$.

Step 2: Rewrite the integral in terms of $u$. Substitute $u$ and $dx$ into the integral: $\int \sin(\sqrt{x})\,dx = \int \sin(u) \cdot 2u\,du = 2 \int u \sin(u)\,du$

Step 3: Apply integration by parts. For the integral $\int u \sin(u)\,du$, use the integration by parts formula $\int v\,dw = vw - \int w\,dv$. Let $v = u$ and $dw = \sin(u)\,du$. Then $dv = du$ and $w = \int \sin(u)\,du = -\cos(u)$.

Step 4: Substitute these into the integration by parts formula. $\int u \sin(u)\,du = u(-\cos(u)) - \int (-\cos(u))\,du$ $= -u\cos(u) + \int \cos(u)\,du$ $= -u\cos(u) + \sin(u)$

Step 5: Combine the results. Now, substitute this back into the expression from Step 2: $2 \int u \sin(u)\,du = 2(-u\cos(u) + \sin(u)) + C$

Step 6: Substitute back $u = \sqrt{x}$. Replace $u$ with $\sqrt{x}$ to express the answer in terms of $x$: $2(-\sqrt{x}\cos(\sqrt{x}) + \sin(\sqrt{x})) + C$ $= 2\sin(\sqrt{x}) - 2\sqrt{x}\cos(\sqrt{x}) + C$

The final answer is $\boxed{2\sin(\sqrt{x}) - 2\sqrt{x}\cos(\sqrt{x}) + C}$.

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