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

∫_( 0) ^(2π) ∣ cos x−sin x ∣dx =

$$\underset{\:\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\:\mid\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\:\mid{dx}\:= \\ $$

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

$${trying}\:{to}\:{solve}.. \\ $$$${see}\:{when}\:\:\frac{\pi}{\mathrm{4}}>{x}\geqslant\mathrm{0}\:\:{cosx}>{sinx}\:\:{so}\:\left({cosx}−{sinx}\right)>\mathrm{0}\:\mid{cosx}−{sinx}\mid=\left({cosx}−{sinx}\right) \\ $$$${when}\:\frac{\pi}{\mathrm{2}}>{x}\geqslant\frac{\pi}{\mathrm{4}}\:{cosx}<{sinx}\:{so}\left({cosx}−{sinx}\right)<\mathrm{0}\:\mid{cosx}−{sinx}\mid=−\left({cosx}−{sinx}\right) \\ $$$${when}\:\pi>{x}\geqslant\frac{\pi}{\mathrm{2}}\:{cosx}\:<\mathrm{0},\:{sinx}>\mathrm{0}\:{so}\left({cosx}−{sinx}\right)<\mathrm{0}\:\mid{cosx}−{sinx}\mid=−\left({cosx}−{sinx}\right) \\ $$$$\frac{\mathrm{5}\pi}{\mathrm{4}}>{x}\geqslant\pi\:\:\mid{cosx}\mid>\mid{sinx}\mid\:{so}\:\left({cosx}−{sinx}\right)<\mathrm{0}\:\mid{cosx}−{sinx}\mid=−\left({cosx}−{sinx}\right) \\ $$$$\frac{\mathrm{6}\pi}{\mathrm{4}}>{x}\geqslant\frac{\mathrm{5}\pi}{\mathrm{4}}\:\:\mid{cosx}\mid<\mid{sinx}\mid\:{so}\:\left({cosx}−{sinx}\right)>\mathrm{0}\:\:{so}\:\mid{cosx}−{sinx}\mid=\left({cosx}−{sinx}\right) \\ $$$$\mathrm{2}\pi>{x}\geqslant\frac{\mathrm{6}\pi}{\mathrm{4}}\:\:{cosx}>\mathrm{0}\:{but}\:{sinx}<\mathrm{0}\:\left({cosx}−{sinx}\right)>\mathrm{0}\:\mid{cosx}−{sinx}\mid=\left({cosx}−{sinx}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left({cosx}−{sinx}\right){dx}+\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} −\:\left({cosx}−{sinx}\right){dx}+ \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:−\left({cosx}−{sinx}\right){dx}+\int_{\pi} ^{\frac{\mathrm{5}\pi}{\mathrm{4}}} \:−\left({cosx}−{sinx}\right){dx} \\ $$$$\int_{\frac{\mathrm{5}\pi}{\mathrm{4}}} ^{\frac{\mathrm{6}\pi}{\mathrm{4}}} \:\left({cosx}−{sinx}\right){dx}+\int_{\frac{\mathrm{6}\pi}{\mathrm{4}}} ^{\mathrm{2}\pi} \:\left({cosx}−{sinx}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid{cosx}−{sinx}\mid{dx}= \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left({cosx}−{sinx}\right){dx}+\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{5}\pi}{\mathrm{4}}} \:−\left({cosx}−{sinx}\right){dx}+\int_{\frac{\mathrm{5}\pi}{\mathrm{4}}} ^{\mathrm{2}\pi} \:\left({cosx}−{sinx}\right){dx} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{check}}\:\boldsymbol{{uoto}}\:\boldsymbol{{this}}\:\boldsymbol{{step}}... \\ $$$$=\mid{sinx}+{cosx}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −\mid{sinx}+{cosx}\mid_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{5}\pi}{\mathrm{4}}} +\mid{sinx}+{cosx}\mid_{\frac{\mathrm{5}\pi}{\mathrm{4}}} ^{\mathrm{2}\pi} \\ $$$$=\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\mathrm{1}\right)−\left\{\left(\frac{−\mathrm{1}}{\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)−\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\right\}+\left\{\left(\mathrm{0}+\mathrm{1}\right)−\left(−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\right\} \\ $$$$=\sqrt{\mathrm{2}}\:−\mathrm{1}+\frac{\mathrm{4}}{\sqrt{\mathrm{2}}}+\mathrm{1}+\frac{\mathrm{2}}{\sqrt{\mathrm{2}}} \\ $$$$=\sqrt{\mathrm{2}}\:+\mathrm{2}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}}\: \\ $$$$=\mathrm{4}\sqrt{\mathrm{2}}\:\:\boldsymbol{{it}}\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{answer}} \\ $$

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