Question

The integral $\int^{\pi/2}_{0}\left|sin\,x-cos\,x\right|dx$ is equal to :

• A. $2\sqrt{2}$
• B. $2\left(\sqrt{2}-1\right)$
• C. $\sqrt{2}+1$
• D. None of these

Answer 1

Answer: (B)

$2\left(\sqrt{2}-1\right)$

Answer 2

We have $cos\,x \ge sin\,x for 0\le x\le\frac{\pi}{4}$ and $sin\,x \ge cos\,x for \frac{\pi}{4}\le x \le\frac{\pi}{2}$ $\therefore \int^{\frac{\pi}{2}}_{0}\left|sin\,x-cos\,x\right|dx$ $=\int^{\frac{\pi}{4}}_{0}\left(cos\,x-sin\,x\right)dx+\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\left(sin\,x-cos\,x\right)dx$ $=\left[sin\,x+cos\,x\right]^{\frac{\pi}{4}}_{0}+\left[-cos\,x-sin\,x\right]^{\frac{\pi}{2}}_{\frac{\pi}{4}}$ $=\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1\right]-\left[0+1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right]$ $\sqrt{2}-1-1+\sqrt{2}=2\sqrt{2}-2$