Question

The value of $\underset{-\frac{\pi }{4}}{\overset{n\pi -\frac{\pi }{4}}{\int }}|\sin x+\cos x|dx$ is:

• A. $\sqrt{2n}$
• B. $4\sqrt{2}n$
• C. $\sqrt{2}n$
• D. $2\sqrt{2}n$

Answer 1

The correct option is D $2\sqrt{2}n$

Let $I=\underset{-\frac{\pi }{4}}{\overset{n\pi -\frac{\pi }{4}}{\int }}|\sin x+\cos x|dx$
$\Rightarrow I=\underset{-\frac{\pi }{4}}{\overset{n\pi -\frac{\pi }{4}}{\int }}\sqrt{2}\left|\sin (x+\frac{\pi }{4})\right|dx$
Taking $x+\frac{\pi }{4}=t\Rightarrow dx=dt$
$\Rightarrow I=\underset{0}{\overset{n\pi }{\int }}\sqrt{2}\left|\sin t\right|dt\Rightarrow I=n\sqrt{2}\underset{0}{\overset{\pi }{\int }}\left|\sin t\right|dt[\because \underset{0}{\overset{nT}{\int }}f\left(x\right)dx=n\underset{0}{\overset{T}{\int }}f\left(x\right)dx]\therefore I=2\sqrt{2}n$