I am given [MATH]\int\sqrt{1-\cos{x}}.[/MATH] I set [MATH]x=2u[/MATH] so that [MATH]dx=2du[/MATH]. Then I get
[MATH] \begin{align*} &2\int\sqrt{1-\cos{2u}}\,du\\ &=2\sqrt{2}\int\sqrt{\frac{1-\cos{2u}}{2}}\,du\\ &=2\sqrt{2}\int\sqrt{\sin^2u}\,du\\ &=2\sqrt{2}\int\sin{u}\,du\\ &=-2\sqrt{2}cos(x/2). \end{align*} [/MATH]It's the same as the answer at the back of the book. My question is, why do we not have plus or minus square root of sin^2 u??? why is it only the positive square root???
[MATH] \begin{align*} &2\int\sqrt{1-\cos{2u}}\,du\\ &=2\sqrt{2}\int\sqrt{\frac{1-\cos{2u}}{2}}\,du\\ &=2\sqrt{2}\int\sqrt{\sin^2u}\,du\\ &=2\sqrt{2}\int\sin{u}\,du\\ &=-2\sqrt{2}cos(x/2). \end{align*} [/MATH]It's the same as the answer at the back of the book. My question is, why do we not have plus or minus square root of sin^2 u??? why is it only the positive square root???