Question

calcule a integral com x variando de 0 a pi de raiz quadrada de 1+cos(X) dx. Como resolver?

Answer 1

$\int\limits_0^\pi{ \sqrt{1+cos(x)} } \, dx$


para resolver vamos multiplicar e dividir por √(1-cos(x) )
 
 $\int\limits_0^\pi{ \sqrt{1+cos(x)} * \frac{\sqrt{1-cos(x)}}{\sqrt{1-cos(x)}} } \, dx\\\\ =\int\limits_0^\pi { \frac{ \sqrt{(1+cos(x))*(1-cos(x))} }{ \sqrt{1-cos(x)} } }\;dx\\\\ =\int\limits_0^\pi { \frac{ \sqrt{1-cos^2(x)} }{ \sqrt{1-cos(x)} } }\;dx\\\\ =\int\limits_0^\pi { \frac{ \sqrt{sen^2(x)} }{ \sqrt{1-cos(x)} } }\;dx\\\\ \boxed{\boxed{ =\int\limits_0^\pi { \frac{sen(x) }{ \sqrt{1-cos(x)} } }\;dx }}$

fazendo a subtituição 
u = 1-cos(x)
du = sen(x) dx  ----> dx = du/sen(x)

$=\int\limits_0^\pi { \frac{sen(x) }{ \sqrt{u} } }*\frac{du}{sen(x)} \\\\ =\int\limits_0^\pi \frac{1}{\sqrt u} \; du = 2\sqrt{u} \left \right | ^\pi_0 =2\sqrt{1-cos(x)} \left \right | ^\pi_0 \\\\\\ = 2 \sqrt{1-cos(\pi) }- 2 \sqrt{1-cos(0) } = \boxed{\boxed{2 \sqrt{2}}}$