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My textbook says I should solve the following integral by first making a substitution, and then using integration by parts:

$$\int \cos\sqrt x \ dx$$

The problem is, after staring at it for a while I'm still not sure what substitution I should make, and hence I'm stuck at the first step. I thought about doing something with the $\sqrt x$, but that doesn't seem to lead anywhere as far as I can tell. Same with the $\cos$. Any hints?

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    $\begingroup$ Don't just think about it, actually try doing it. The substitution $t = \sqrt{x}$ is correct. $\endgroup$ Commented Mar 6, 2015 at 3:59

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make a subs $u = \sqrt x, x = u^2, dx = 2u du$ now the integral $\int \cos \sqrt x \, dx$ is transformed into $$2\int u \cos u \, du = 2 \int u d (\sin u) =2\left( u\sin u - \int \sin u \, du\right) = 2\left( u\sin u +\cos u +C\right)$$

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  • $\begingroup$ That's quite a hint. :) $\endgroup$ Commented Mar 3, 2015 at 4:06
  • $\begingroup$ @Asker, that was an error. fixed it. $\endgroup$ Commented Mar 3, 2015 at 4:29
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Try $x = u^2$, and $dx = 2u ~ du$.

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Let $x=t^2$, then $dx=2tdt$, so $$\int\cos\sqrt{x}dx=\int 2t\cos tdt$$

Then use integration by parts.

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