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I need help finding the integral of $\sin(\sqrt{x})dx$. I have the answer here but would like to know how to get there.

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Hint: Use the substitution $\sqrt{x}=u$ or $x=u^2$ and then integration by parts.

$$\int \sin(\sqrt{x})dx = 2\int u\sin(u)du $$

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Use the substitution $t=\sqrt{x}$, and then use integration by parts. From the substitution, you should get:

$$\int\sin\sqrt{x}dx=2\int t\sin (t) dt$$

Then, use integration by parts. Let $u=t$ and $dv=\sin t$. Then $du=dt$ and $v=-\cos t$, and we have:

$$\int t\sin (t) dt=-t\cos t +\int \cos (t) dt=\sin t -t\cos t=\sin\sqrt{x}-\sqrt{x}\cos\sqrt{x}$$

and we have the solution:

$$\int\sin\sqrt{x}dx=2(\sin\sqrt{x}-\sqrt{x}\cos\sqrt{x})$$

I hope this clears up the matter for you!

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Thank you so much! –  Jaden M. Jan 6 '13 at 5:20
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$$J = \underbrace{\int \sin(\sqrt{x})dx = \int \sin(t) 2t dt }_{\sqrt{x} = t \implies x = t^2 \implies dx = 2t dt}$$ $$I = \int t \sin(t) dt = -\int t d(\cos(t)) = - \left(t \cos(t) - \int \cos(t) dt \right) = - t \cos(t) + \sin(t) + c$$ Hence, $$J = - 2t \cos(t) + 2\sin(t) + k = -2 \sqrt{x} \cos(\sqrt{x}) + 2 \sin(\sqrt{x}) + k$$

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Hint: $$\sin(\sqrt{x}) = \frac{\sqrt{x}\sin(\sqrt{x})}{\sqrt{x}}$$ and take $u=\sqrt{x}$, $2 du=\frac{dx}{\sqrt{x}}$.

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