∫_( 0) ^(2π) ∣ cos x−sin x ∣dx =


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Question Number 58211 by salaw2000 last updated on 20/Apr/19

$\underset{0}{\int^{2 π}} ∣ \cos x - \sin x ∣ d x =$
Commented by tanmay last updated on 20/Apr/19

Commented by tanmay last updated on 20/Apr/19
$trying to solve.. see when (π/4)>x≥0 cosx>sinx so (cosx−sinx)>0 ∣cosx−sinx∣=(cosx−sinx) when (π/2)>x≥(π/4) cosx<sinx so(cosx−sinx)<0 ∣cosx−sinx∣=−(cosx−sinx) when π>x≥(π/2) cosx <0, sinx>0 so(cosx−sinx)<0 ∣cosx−sinx∣=−(cosx−sinx) ((5π)/4)>x≥π ∣cosx∣>∣sinx∣ so (cosx−sinx)<0 ∣cosx−sinx∣=−(cosx−sinx) ((6π)/4)>x≥((5π)/4) ∣cosx∣<∣sinx∣ so (cosx−sinx)>0 so ∣cosx−sinx∣=(cosx−sinx) 2π>x≥((6π)/4) cosx>0 but sinx<0 (cosx−sinx)>0 ∣cosx−sinx∣=(cosx−sinx) ∫_0 ^(π/4) (cosx−sinx)dx+∫_(π/4) ^(π/2) − (cosx−sinx)dx+ ∫_(π/2) ^π −(cosx−sinx)dx+∫_π ^((5π)/4) −(cosx−sinx)dx ∫_((5π)/4) ^((6π)/4) (cosx−sinx)dx+∫_((6π)/4) ^(2π) (cosx−sinx)dx ∫_0 ^(2π) ∣cosx−sinx∣dx= =∫_0 ^(π/4) (cosx−sinx)dx+∫_(π/4) ^((5π)/4) −(cosx−sinx)dx+∫_((5π)/4) ^(2π) (cosx−sinx)dx pls check uoto this step... =∣sinx+cosx∣_0 ^(π/4) −∣sinx+cosx∣_(π/4) ^((5π)/4) +∣sinx+cosx∣_((5π)/4) ^(2π) =((1/(√2))+(1/(√2))−1)−{(((−1)/(√2))−(1/(√2)))−((1/(√2))+(1/(√2)))}+{(0+1)−(−(1/(√2))−(1/(√2)))} =(√2) −1+(4/(√2))+1+(2/(√2)) =(√2) +2(√2) +(√2) =4(√2) it is the answer$
$t r y i n g t o s o l v e . .$ $s e e w h e n \frac{π}{4} > x ⩾ 0 c o s x > s i n x s o (c o s x - s i n x) > 0 ∣ c o s x - s i n x ∣= (c o s x - s i n x)$ $w h e n \frac{π}{2} > x ⩾ \frac{π}{4} c o s x < s i n x s o (c o s x - s i n x) < 0 ∣ c o s x - s i n x ∣= - (c o s x - s i n x)$ $w h e n π > x ⩾ \frac{π}{2} c o s x < 0, s i n x > 0 s o (c o s x - s i n x) < 0 ∣ c o s x - s i n x ∣= - (c o s x - s i n x)$ $\frac{5 π}{4} > x ⩾ π ∣ c o s x ∣>∣ s i n x ∣ s o (c o s x - s i n x) < 0 ∣ c o s x - s i n x ∣= - (c o s x - s i n x)$ $\frac{6 π}{4} > x ⩾ \frac{5 π}{4} ∣ c o s x ∣<∣ s i n x ∣ s o (c o s x - s i n x) > 0 s o ∣ c o s x - s i n x ∣= (c o s x - s i n x)$ $2 π > x ⩾ \frac{6 π}{4} c o s x > 0 b u t s i n x < 0 (c o s x - s i n x) > 0 ∣ c o s x - s i n x ∣= (c o s x - s i n x)$ $\int_{0}^{\frac{π}{4}} (c o s x - s i n x) d x + \int_{\frac{π}{4}}^{\frac{π}{2}} - (c o s x - s i n x) d x +$ $\int_{\frac{π}{2}}^{π} - (c o s x - s i n x) d x + \int_{π}^{\frac{5 π}{4}} - (c o s x - s i n x) d x$ $\int_{\frac{5 π}{4}}^{\frac{6 π}{4}} (c o s x - s i n x) d x + \int_{\frac{6 π}{4}}^{2 π} (c o s x - s i n x) d x$ $\int_{0}^{2 π} ∣ c o s x - s i n x ∣ d x =$ $= \int_{0}^{\frac{π}{4}} (c o s x - s i n x) d x + \int_{\frac{π}{4}}^{\frac{5 π}{4}} - (c o s x - s i n x) d x + \int_{\frac{5 π}{4}}^{2 π} (c o s x - s i n x) d x$ $p l s c h e c k u o t o t h i s s t e p . . .$ $=∣ s i n x + c o s x ∣_{0}^{\frac{π}{4}} - ∣ s i n x + c o s x ∣_{\frac{π}{4}}^{\frac{5 π}{4}} + ∣ s i n x + c o s x ∣_{\frac{5 π}{4}}^{2 π}$ $= (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1) - {(\frac{- 1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) - (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})} + {(0 + 1) - (- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})}$ $= \sqrt{2} - 1 + \frac{4}{\sqrt{2}} + 1 + \frac{2}{\sqrt{2}}$ $= \sqrt{2} + 2 \sqrt{2} + \sqrt{2}$ $= 4 \sqrt{2} i t i s t h e a n s w e r$
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