((bobhans)/(⋰⋱)) ∫_(π/2) ^π ∣ cos x−sin x ∣ dx ?


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Question Number 108382 by bobhans last updated on 16/Aug/20

$\frac{b o b h a n s}{\iddots ⋱}$ $\underset{π / 2}{\int^{π}} ∣ \cos x - \sin x ∣ d x ?$
Commented by PRITHWISH SEN 2 last updated on 16/Aug/20

$by shifting the entire function by π we get$ $\int_{\frac{3 π}{2}}^{2 π} (\cos x - \sin x) dx = \sin x + \cos x ∣_{\frac{3 π}{2}}^{2 π} = 2$
	Answered by bemath last updated on 16/Aug/20

	$\frac{△ \frac{B e}{M a t h} △}{△}$ $w e k n o w t h a t \cos x - \sin x < 0$ $f o r \frac{π}{2} < x < π$ $t h e n \underset{π / 2}{\int^{π}} ∣ \cos x - \sin x ∣ d x =$ ${\underset{π / 2}{\int^{π}} (\sin x - \cos x) d x = (- \cos x - \sin x)]}_{\frac{π}{2}}^{π}$ $= 1 - (- 1) = 2$
	Answered by 1549442205PVT last updated on 17/Aug/20

	$cosx - sinx = \sqrt{2} (\frac{\sqrt{2}}{2} cosx - \frac{\sqrt{2}}{2} sinx)$ $= \sqrt{2} (\sin \frac{π}{4} cosx - \cos \frac{π}{4} sinx)$ $= \sqrt{2} \sin (\frac{π}{4} - x) = - \sqrt{2} . \sin (x - \frac{π}{4}) < 0$ $for x \in [\frac{π}{2}; π] . Hence, I = \underset{π / 2}{\int^{π}} ∣ \cos x - \sin x ∣ d x$ $= \underset{π / 2}{\int^{π}} (sinx - \cos x) d x$ $= (- cosx - sinx) ∣_{\frac{π}{2}}^{π} = (1 - 0) - (0 - 1)$ $= 2$ $or I = \sqrt{2} \int \sin (x - \frac{π}{4}) dx = - \sqrt{2} \cos (x - \frac{π}{4}) ∣_{\frac{π}{2}}^{π}$ $= - \sqrt{2} (\cos \frac{3 π}{4} - \cos \frac{π}{4}) = - \sqrt{2} (- \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2})$ $= - \sqrt{2} \times (- \sqrt{2}) = 2$
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