<s>[INST] $\int \cos^5 x \sin^{1/2} x \ dx $ $$I=\int \cos^5 x \sin^{1/2} x \ dx $$
My attempt:
$$\cos^5 x = \cos^4 x \cos x =(1-\sin^2x)^2 \cos x$$
\begin{align}
I & =\int (1-\sin^2x)^2 \sin^{1/2} x \cos x \ dx \\[10pt]
& =\int \sin^{1/2} x \cos x \ dx+\int \sin^2 x \cos x \, dx + \int -2\sin x \cos x\, dx \\[10pt]
& ={2\over 3} \sin^{3\over 2} x+{1\over 3} \sin^3x+\cos^2 x+c
\end{align}
True ?
 [/INST] You made a tiny mistake with the powers of $\,sin(x)$
$I=\int \cos^5 x \sin^{1/2} x \ dx$
$I  =\int (1-\sin^2x)^2 \sin^{1/2} x \cos x \ dx $
let
 $\,u= sinx  \implies du = cos(x)dx$
$I= \int (1-u^2)^2.u^\frac12\,du \\I = \int u^\frac12 + u^{4+\frac12} - 2u^{2+\frac12}\,du\\I = \int u^\frac12+ u^\frac92 - 2u^\frac52\,du\\I = \frac23u^\frac32 + \frac{2}{11}u^\frac{11}{2}-\frac47u^\frac72 +C \\I= \frac23(sin(x))^\frac32+\frac{2}{11}(sin(x))^\frac{11}{2}- \frac47(sin(x))^\frac74 +C$
</s>