# How does &#x222B;<!-- ∫ --> cos 4 </msup> &#x2

How does $\int \frac{{\mathrm{cos}}^{4}x}{\sqrt{1-\mathrm{sin}x}}\phantom{\rule{thinmathspace}{0ex}}dx$ simplify to $-\int -{\left(2-u\right)}^{\frac{3}{2}}u\phantom{\rule{thinmathspace}{0ex}}du$
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dariajoq9
${\mathrm{cos}}^{3}x={\left(\sqrt{1-{\mathrm{sin}}^{2}x}\right)}^{3}={\left(\sqrt{1-\left(1-u{\right)}^{2}}\right)}^{3}={\left(\sqrt{u\left(2-u\right)}\right)}^{3}=u\sqrt{u}\left(2-u{\right)}^{\frac{3}{2}}$