Question

\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]

Answer 1

\[\int_0^\frac{\pi}{2} \left| \sin x - \cos x \right| d x\]
\[ = \sqrt{2} \int_0^\frac{\pi}{2} \left| \sin x\frac{1}{\sqrt{2}} - \cos x\frac{1}{\sqrt{2}} \right| d x\]
\[ = \sqrt{2} \int_0^\frac{\pi}{2} \left| \sin x \cos\frac{\pi}{4} - \cos x \sin\frac{\pi}{4} \right| d x\]
\[ = \sqrt{2} \int_0^\frac{\pi}{2} \left| \sin\left( x - \frac{\pi}{4} \right) \right| d x\]
\[We have, \]
\[\left| \sin\left( x - \frac{\pi}{4} \right) \right| = \begin{cases} - \sin\left( x - \frac{\pi}{4} \right),& 0 \leq x \leq \frac{\pi}{4}\\ \sin\left( x - \frac{\pi}{4} \right),& \frac{\pi}{4} \leq x \leq \frac{\pi}{2}\end{cases}\]
\[ \therefore \int_0^\frac{\pi}{2} \left| \sin x - \cos x \right| d x = \sqrt{2} \int_0^\frac{\pi}{4} - \sin\left( x - \frac{\pi}{4} \right) d x + \sqrt{2} \int_\frac{\pi}{4}^\frac{\pi}{2} \sin\left( x - \frac{\pi}{4} \right) d x\]
\[ = \sqrt{2} \left[ \cos\left( x - \frac{\pi}{4} \right) \right]_0^\frac{\pi}{4} - \sqrt{2} \left[ \cos\left( x - \frac{\pi}{4} \right) \right]_\frac{\pi}{4}^\frac{\pi}{2} \]
\[ = \sqrt{2}\left[ \cos \left( 0 \right) - \cos\left( - \frac{\pi}{4} \right) \right] - \sqrt{2}\left[ \cos\left( \frac{\pi}{4} \right) - \cos \left( 0 \right) \right]\]
\[ = \sqrt{2}\left( 1 - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + 1 \right)\]
\[ = \sqrt{2}\left( 2 - \frac{2}{\sqrt{2}} \right)\]
\[ = 2\sqrt{2} - 2\]
\[ = 2\left( \sqrt{2} - 1 \right)\]