Question

Area bounded by the curve $y=1$ and $y=\frac{\sin x+\cos x+|\sin x-\cos x|}{2}\text{in}x\in [0,\pi ]\text{is}a\pi -\sqrt{b}-c,$ (where $a,b,c$ are integers) then $a+b+c$ is

Answer 1

$\frac{\sin x+\cos x+|\sin x-\cos x|}{2}=\{\begin{array}{c}\cos x,x\in (0,\frac{\pi }{4})\\ \sin x,x\in (\frac{\pi }{4},\pi )\end{array}$

So,required area=$\underset{0}{\overset{\frac{\pi }{4}}{\int }}(1-\cos x)dx+\underset{\frac{\pi }{4}}{\overset{\pi }{\int }}(1-\sin x)dx$
$=\pi -\sqrt{2}-1\Rightarrow a+b+c=4$