Question

\[\int_\pi^\frac{3\pi}{2} \sqrt{1 - \cos2x}dx\]

Answer 1

\[\int_\pi^\frac{3\pi}{2} \sqrt{1 - \cos2x}dx\]

\[= \int_\pi^\frac{3\pi}{2} \sqrt{2 \sin^2 x}dx\]
\[ = \sqrt{2} \int_\pi^\frac{3\pi}{2} \left| \sin x \right|dx\]
\[ = - \sqrt{2} \int_\pi^\frac{3\pi}{2} \sin x\ dx .................\left( \sin x < 0 for\ \pi \leq x \leq 2\pi \right)\]

\[= - \sqrt{2}\left( - \cos x \right) |_\pi^\frac{3\pi}{2} \]
\[ = \sqrt{2}\left( \cos\frac{3\pi}{2} - cos\pi \right)\]
\[ = \sqrt{2} \left[ 0 - \left( - 1 \right) \right]\]
\[ = \sqrt{2} \times 1\]
\[ = \sqrt{2}\]