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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

I got stuck on integrate this function

\(\displaystyle \int \frac{\sin^3(\sqrt{x})}{\sqrt{x}}dx\)

my first thinking was rewrite it as \(\displaystyle \int \frac{\sin^2(\sqrt{x})\sin(\sqrt{x})}{\sqrt{x}}dx\)

then use the identity \(\displaystyle \cos^2(x)+\sin^2(x)=1 \ \therefore \sin^2x=1- \cos^2(x)\)

\(\displaystyle \int \frac{(1-\cos^2(\sqrt{x}))\sin(\sqrt{x})}{\sqrt{x}}dx\)

subsitute \(\displaystyle u= \cos(x) \therefore du=- \sin(x) dx\)

then we get

\(\displaystyle - \int \frac{1-u^2}{ \cos^{-1}(u)}du\)

but that does not seem smart so my last ide is integrate by part, but I struggle on that part..

\(\displaystyle u= \sqrt{x}^{-1} \therefore du=\sqrt{x}^{-2}\) and \(\displaystyle dv=\sin^3(\sqrt{x}) \therefore v=?\)

Regards,

\(\displaystyle |\pi\rangle\)