Question

I don't know how to integrate this anti derivative

Answer 1

`\bf \int\limits_{(\pi )/(4)}^{(\pi )/(2)}\cfrac{2cos(x)}{√(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_{(\pi )/(4)}^{(\pi )/(2)}\cfrac{2cos(x)}{√(u)}\cdot \cfrac{du}{cos(x)}\implies \int\limits_{(\pi )/(4)}^{(\pi )/(2)}\cfrac{2}{√(u)}\cdot du\\ -----------------------------\\`


`\bf \textit{now, let us do the bounds} \\\\ u\left( (\pi )/(2) \right)=1+sin\left( (\pi )/(2) \right)\to 2 \\\\ u\left( (\pi )/(4) \right)=1+sin\left( (\pi )/(4) \right)\to 1+\cfrac{√(2)}{2}\to \cfrac{2+√(2)}{2}\\ -----------------------------\\ thus\implies \int\limits_{(\pi )/(4)}^{(\pi )/(2)}\cfrac{2}{√(u)}\cdot du\implies \int\limits_{(2+√(2) )/(2)}^(2)\cfrac{2}{√(u)}\cdot du \\\\`


`\bf 2\int\limits_{(2+√(2) )/(2)}^(2) u^{-(1)/(2)}\implies 2\cdot \cfrac{u^{(1)/(2)}}{(1)/(2)}\implies \left[\cfrac{}{} 4√(u) \right]_{(2+√(2) )/(2)}^(2)`