<s>[INST] How to integrate $\int \frac {\cos^4 x}{\sqrt {1- \sin x}} $ I frankly don't know where to start. Looked for some identities to get rid of the square root below.
$$\int \frac {\cos^4 x}{\sqrt {1- \sin x}} \, dx  $$
 [/INST] Write $\cos(x)^4 = \cos(x)(1-\sin^2(x))^{3/2} = \cos(x)(1-\sin(x))^{3/2}(1+\sin(x))^{3/2}.$ Thus $$\int \frac{\cos(x)^4}{\sqrt{1-\sin(x)}}dx = \int (1-\sin(x))(1+\sin(x))^{3/2}\cos(x) dx$$ Put $u = \sin(x)$. Then $$\int\frac{\cos(x)^4}{\sqrt{1-\sin(x)}}dx = \int(1-u)(1+u)^{3/2}du.$$ Now put $z = 1+u$ so $$\int\frac{\cos(x)^4}{\sqrt{1-\sin(x)}}dx = \int(2-z)z^{3/2}dz = \int (2z^{3/2} - z^{5/2}) dz = \frac{4}{5} z^{5/2} - \frac{2}{7} z^{7/2} + C.$$ Un-doing the substitutions gives $$\int\frac{\cos(x)^4}{\sqrt{1-\sin(x)}}dx =\frac{4}{5}(1+\sin(x))^{5/2} - \frac{2}{7}(1+\sin(x))^{7/2} + C.$$ 
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