Question

The value of integral $\int |\sin x-\cos x|dx$ will be

• A. $\sin x+\cos x+C$ for $x\in [\frac{\pi }{4},\frac{\pi }{2}]$
• B. $\cos x-\sin x+C$ for $x\in [\frac{\pi }{2},\pi ]$
• C. $\sin x+\cos x+C$ for $x\in [0,\frac{\pi }{4}]$
• D. $-(\sin x+\cos x)+C$ for $x\in [\frac{\pi }{4},\frac{\pi }{2}]$

Answer 1

Answer: (C), (D)

$-(\sin x+\cos x)+C$ for $x\in [\frac{\pi }{4},\frac{\pi }{2}]$
$\sin x+\cos x+C$ for $x\in [0,\frac{\pi }{4}]$

We know that $\sin x>\cos x$ in the range $x\in [\frac{\pi }{4},\frac{\pi }{2}]$

And $\cos x>\sin x$ when $x\in [0,\frac{\pi }{4}]$

$\therefore \int |\sin x-\cos x|dx$ becomes $\int (\sin x-\cos x)dx$ when $x\in [\frac{\pi }{4},\frac{\pi }{2}]$

i.e. $-\cos x-\sin x+c$ for $x\in [\frac{\pi }{4},\frac{\pi }{2}]$

The integral can be written as $\int (\cos x-\sin x)dx$ for $x\in [0,\frac{\pi }{4}]$

i.e. $\sin x+\cos x+c$ for $x\in [0,\frac{\pi }{4}]$