<s>[INST] Integrate $\int\frac{\cos^5(x)}{\sin(x)^{1/2}}dx$ I can't find a way to solve this:
$$\int\frac{\cos^5(x)}{\sin(x)^{1/2}}dx$$
I'm not sure how i have to proceed, can you help me? I've tried the substitution u=(sinx), but it got me nowhere.
 [/INST] Hint: 
(Long) METHOD 1: See that
$$\dfrac{d\sqrt{\sin x}}{dx}=\dfrac{1}{2\sqrt{\sin x}}\cos x\\
\implies \dfrac{1}{\sqrt{\sin x}}\cos xdx=2d\sqrt{\sin x}$$
Thus,
$$\int\dfrac{\cos^5x}{\sqrt{\sin x}}dx=2\int{\cos^4x}d\sqrt{\sin x}=\\
2\left(\sqrt{\sin x}\cos^4x+4\int\sqrt{\sin x}\cos^3 x\sin xdx\right)$$
Now, $\int\cos^3 x\sin^{3/2} xdx$ can be calculated by $\dfrac{d\sin^{5/2}x}{dx}=\dfrac{5}{2}\sin^{3/2}x\cos x$, i.e., $\sin^{3/2}x\cos xdx=\dfrac{2}{5}d\sin^{5/2}x$. Continue integration by parts.
(Shorter) METHOD 2: Use the identity $1-\sin^2 =\cos^2$, to get
$$\int\dfrac{(1-\sin^2x)^2}{\sqrt{\sin x}}dx=2\int2{(1-u^4)^2}du,\text{ $u=\sqrt{\sin x}$}\\
\implies \int\dfrac{(1-\sin^2x)^2}{\sqrt{\sin x}}dx=\int(1+u^8-2u^4)du=\\
\boxed{\sqrt{\sin x}+\dfrac{\sqrt{\sin^9 x}}{9}-2\dfrac{\sqrt{\sin^5x}}{5}+C}$$
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