Petrus
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Hello MHB,
I got stuck on integrate this function
$$\int \frac{\sin^3(\sqrt{x})}{\sqrt{x}}dx$$
my first thinking was rewrite it as $$\int \frac{\sin^2(\sqrt{x})\sin(\sqrt{x})}{\sqrt{x}}dx$$
then use the identity $$\cos^2(x)+\sin^2(x)=1 \ \therefore \sin^2x=1- \cos^2(x)$$
$$\int \frac{(1-\cos^2(\sqrt{x}))\sin(\sqrt{x})}{\sqrt{x}}dx$$
subsitute $$u= \cos(x) \therefore du=- \sin(x) dx$$
then we get
$$- \int \frac{1-u^2}{ \cos^{-1}(u)}du$$
but that does not seem smart so my last ide is integrate by part, but I struggle on that part..
$$u= \sqrt{x}^{-1} \therefore du=\sqrt{x}^{-2}$$ and $$dv=\sin^3(\sqrt{x}) \therefore v=?$$
Regards,
$$|\pi\rangle$$
I got stuck on integrate this function
$$\int \frac{\sin^3(\sqrt{x})}{\sqrt{x}}dx$$
my first thinking was rewrite it as $$\int \frac{\sin^2(\sqrt{x})\sin(\sqrt{x})}{\sqrt{x}}dx$$
then use the identity $$\cos^2(x)+\sin^2(x)=1 \ \therefore \sin^2x=1- \cos^2(x)$$
$$\int \frac{(1-\cos^2(\sqrt{x}))\sin(\sqrt{x})}{\sqrt{x}}dx$$
subsitute $$u= \cos(x) \therefore du=- \sin(x) dx$$
then we get
$$- \int \frac{1-u^2}{ \cos^{-1}(u)}du$$
but that does not seem smart so my last ide is integrate by part, but I struggle on that part..
$$u= \sqrt{x}^{-1} \therefore du=\sqrt{x}^{-2}$$ and $$dv=\sin^3(\sqrt{x}) \therefore v=?$$
Regards,
$$|\pi\rangle$$