# How do you integrate int x sin sqrtx dx  using integration by parts?

Jan 24, 2016

let's $u = \sqrt{x}$

$\mathrm{du} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$

$2 u \mathrm{du} = \mathrm{dx}$

${u}^{2} = x$

so you have

$2 \int {u}^{3} \sin \left(u\right) \mathrm{du}$

And integrate by part three time and you have

$2 \left[\left(3 \left({u}^{2} - 2\right) \sin \left(u\right) - u \left({u}^{2} - 6\right) \cos \left(u\right)\right)\right] + C$