[tex]\bf \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}\cfrac{2cos(x)}{\sqrt{1+sin(x)}}\cdot \underline{dx}\\
-----------------------------\\
u=1+sin(x)
\\\\
\cfrac{du}{dx}=cos(x)\implies \cfrac{du}{cos(x)}=\underline{dx}\\
-----------------------------\\
thus\implies \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}\cfrac{2cos(x)}{\sqrt{u}}\cdot \cfrac{du}{cos(x)}\implies
\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}\cfrac{2}{\sqrt{u}}\cdot du\\
-----------------------------\\[/tex]
[tex]\bf \textit{now, let us do the bounds}
\\\\
u\left( \frac{\pi }{2} \right)=1+sin\left( \frac{\pi }{2} \right)\to 2
\\\\
u\left( \frac{\pi }{4} \right)=1+sin\left( \frac{\pi }{4} \right)\to 1+\cfrac{\sqrt{2}}{2}\to \cfrac{2+\sqrt{2}}{2}\\
-----------------------------\\
thus\implies \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}}\cfrac{2}{\sqrt{u}}\cdot du\implies
\int\limits_{\frac{2+\sqrt{2} }{2}}^{2}\cfrac{2}{\sqrt{u}}\cdot du
\\\\
[/tex]
[tex]\bf 2\int\limits_{\frac{2+\sqrt{2} }{2}}^{2} u^{-\frac{1}{2}}\implies 2\cdot \cfrac{u^{\frac{1}{2}}}{\frac{1}{2}}\implies \left[\cfrac{}{} 4\sqrt{u} \right]_{\frac{2+\sqrt{2} }{2}}^{2}[/tex]
