- #1
Petrus
- 702
- 0
Hello MHB,
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,
\(\displaystyle |\pi\rangle\)
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,
\(\displaystyle |\pi\rangle\)