Discussion Forum

 The integral   $\int_{0}^{\pi} \sqrt{1+4\sin^{2}\frac{x}{2}-4\sin\frac{x}{2}}dx$   is equal to 

Answer:

Explanation:

 Use the formula 

   $|x-a|=\begin{cases}x-a, & x \geq a\\-(x-a), & x < a\end{cases}$

 to break given integral in two parts and then integrate separetely

= $\int_{0}^{\pi} \sqrt{\left(1-2\sin\frac{x}{2}\right)^{2}}dx=\int_{0}^{\pi} |1-2\sin\frac{x}{2}| dx$

=$\int_{0}^{\pi/3} \left(1-2\sin\frac{x}{2}\right)^{}dx=\int_{\pi/3}^{\pi} \left(1-2\sin\frac{x}{2}\right) dx$

=$\left(x+4\cos\frac{x}{2}\right)_{0}^{\pi/3}-\left(x+4\cos\frac{x}{2}\right)_{\pi/3}^{\pi}$

   = $4\sqrt{3}-4-\frac{\pi}{3}$