Question

The value of $\underset{0}{\overset{\pi }{\int }}|\sin x+\cos x|dx$ is

• A. $2$
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. $0$

Answer 1

The correct option is C $2\sqrt{2}$

Let $I=\underset{0}{\overset{\pi }{\int }}\mid \sin x+\cos x|dx$

$\Rightarrow I=\underset{0}{\overset{\pi }{\int }}\sqrt{2}\times \frac{1}{\sqrt{2}}|\sin x+\cos x|dx$

$\Rightarrow I=\sqrt{2}\underset{0}{\overset{\pi }{\int }}|\sin x\cos \frac{\pi }{4}+\cos x\sin \frac{\pi }{4}|dx$

$\Rightarrow I=\sqrt{2}\underset{0}{\overset{\pi }{\int }}\left|\sin (x+\frac{\pi }{4})\right|dx$

$\Rightarrow I=\sqrt{2}\underset{0}{\overset{\frac{3\pi }{4}}{\int }}\sin (x+\frac{\pi }{4})dx-\sqrt{2}\underset{3\pi /4}{\overset{\pi }{\int }}\sin (x+\frac{\pi }{4})dx$
$\Rightarrow {I=\sqrt{2}\mid -\cos (x+\frac{\pi }{4})|}_0^{3\pi /4}+\sqrt{2}{\cos (x+\frac{\pi }{4})|}_{3\pi /4}^{\pi }$

$\Rightarrow I=\sqrt{2}(-(-1-\frac{1}{\sqrt{2}})+(-\frac{1}{\sqrt{2}}+1))$

$\Rightarrow I=\sqrt{2}(1+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+1)$

$\Rightarrow I=2\sqrt{2}$