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Question Number 207352 by NasaSara last updated on 12/May/24

calculate:   ∫_(Π/4) ^(Π/2) ⌊cot(x)⌋ dx

$${calculate}: \\ $$$$\:\int_{\frac{\Pi}{\mathrm{4}}} ^{\frac{\Pi}{\mathrm{2}}} \lfloor{cot}\left({x}\right)\rfloor\:{dx} \\ $$

Commented by NasaSara last updated on 12/May/24

thank you

$${thank}\:{you} \\ $$

Commented by Berbere last updated on 12/May/24

withe Pleasur

$${withe}\:{Pleasur} \\ $$

Commented by Berbere last updated on 12/May/24

0≤cot(x)<1 ;∀x∈](π/4);(π/2)]⇒[cot(x)]=0  ∫_(π/4) ^(π/2) [cot(x)]dx=0

$$\left.\mathrm{0}\left.\leqslant{cot}\left({x}\right)<\mathrm{1}\:;\forall{x}\in\right]\frac{\pi}{\mathrm{4}};\frac{\pi}{\mathrm{2}}\right]\Rightarrow\left[{cot}\left({x}\right)\right]=\mathrm{0} \\ $$$$\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \left[{cot}\left({x}\right)\right]{dx}=\mathrm{0} \\ $$

Answered by MM42 last updated on 12/May/24

(π/4)<x<(π/2)⇒0<cotx<1⇒⌊cotx⌋=0  ⇒I=0 ✓

$$\frac{\pi}{\mathrm{4}}<{x}<\frac{\pi}{\mathrm{2}}\Rightarrow\mathrm{0}<{cotx}<\mathrm{1}\Rightarrow\lfloor{cotx}\rfloor=\mathrm{0} \\ $$$$\Rightarrow{I}=\mathrm{0}\:\checkmark \\ $$

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