# Use the identities cos &#x2061;<!-- ⁡ --> 2 x = 2 cos 2 </msup> &#x2061

Use the identities
$\mathrm{cos}2x=2{\mathrm{cos}}^{2}x-1=1-2{\mathrm{sin}}^{2}x$
$\mathrm{sin}x=\mathrm{cos}\left(\frac{\pi }{2}-x\right)$
to help evaluate
${\int }_{0}^{\pi /2}\sqrt{1-\mathrm{sin}x}\phantom{\rule{thickmathspace}{0ex}}dx$
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odczepneyv
So, if $\mathrm{sin}x=\mathrm{cos}\left(\frac{\pi }{2}-x\right)$, then
${\int }_{x=0}^{\pi /2}\sqrt{1-\mathrm{sin}x}\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{x=0}^{\pi /2}\sqrt{1-\mathrm{cos}\left(\frac{\pi }{2}-x\right)}\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{u=0}^{\pi /2}\sqrt{1-\mathrm{cos}u}\phantom{\rule{thinmathspace}{0ex}}du,$
the last equality due to the substitution $u=\pi /2-x$. Now use the same method of solution, namely $\sqrt{1-\mathrm{cos}u}=\sqrt{2}\mathrm{sin}\frac{u}{2}$, to deduce the result.
###### Not exactly what you’re looking for?
Zeihergp
${\int }_{0}^{\pi /2}\sqrt{1-\mathrm{sin}x}\phantom{\rule{thinmathspace}{0ex}}dx$
$=2{\int }_{0}^{\pi /2}\sqrt{1-\mathrm{sin}x}\phantom{\rule{thinmathspace}{0ex}}d\left(x/2\right)$
$=2{\int }_{0}^{\pi /2}\sqrt{\left(\mathrm{sin}x/2-\mathrm{cos}x/2{\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}d\left(x/2\right)$
$=2{\int }_{0}^{\pi /2}\left(\mathrm{sin}x/2-\mathrm{cos}x/2\right)\phantom{\rule{thinmathspace}{0ex}}d\left(x/2\right)$
that you can further on process taking $x/2=u$ for substitution.