Question-138580

Question Number 138580 by Bekzod Jumayev last updated on 15/Apr/21

Commented by Bekzod Jumayev last updated on 15/Apr/21

$${Please}\:{help}? \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 15/Apr/21

∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx ∫_(−π) ^π ((sin(2021x))/((1+3^x )sinx))dx=∫_(−π) ^π ((sin(2021x))/((1+3^(−x) )sinx))dx=∫_(−π) ^π ((3^x sin(2021x))/((1+3^x )sinx))dx =(1/2)∫_(−π) ^π (((1+3^x )sin(2021x))/((1+3^x )sinx))dx=(1/2)∫_(−π) ^π ((sin(2021x))/(sinx))dx =∫_0 ^π ((sin(2021x))/(sinx))dx

$$\int_{\mathrm{a}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{a}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{a}+\mathrm{b}−\mathrm{x}\right)\mathrm{dx} \\ $$$$\int_{−\pi} ^{\pi} \frac{\mathrm{sin}\left(\mathrm{2021x}\right)}{\left(\mathrm{1}+\mathrm{3}^{\mathrm{x}} \right)\mathrm{sinx}}\mathrm{dx}=\int_{−\pi} ^{\pi} \frac{\mathrm{sin}\left(\mathrm{2021x}\right)}{\left(\mathrm{1}+\mathrm{3}^{−\mathrm{x}} \right)\mathrm{sinx}}\mathrm{dx}=\int_{−\pi} ^{\pi} \frac{\mathrm{3}^{\mathrm{x}} \mathrm{sin}\left(\mathrm{2021x}\right)}{\left(\mathrm{1}+\mathrm{3}^{\mathrm{x}} \right)\mathrm{sinx}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi} ^{\pi} \frac{\left(\mathrm{1}+\mathrm{3}^{\mathrm{x}} \right)\mathrm{sin}\left(\mathrm{2021x}\right)}{\left(\mathrm{1}+\mathrm{3}^{\mathrm{x}} \right)\mathrm{sinx}}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\pi} ^{\pi} \frac{\mathrm{sin}\left(\mathrm{2021x}\right)}{\mathrm{sinx}}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\left(\mathrm{2021x}\right)}{\mathrm{sinx}}\mathrm{dx} \\ $$

Answered by phanphuoc last updated on 15/Apr/21

I=∫_(−π) ^π 3^x sin(2021x)dx/(1+3^x )sinx with x=−u 2I=∫_(−π) ^π sin(2021x)dx/sinx

$${I}=\int_{−\pi} ^{\pi} \mathrm{3}^{{x}} {sin}\left(\mathrm{2021}{x}\right){dx}/\left(\mathrm{1}+\mathrm{3}^{{x}} \right){sinx}\:{with}\:{x}=−{u} \\ $$$$\mathrm{2}{I}=\int_{−\pi} ^{\pi} {sin}\left(\mathrm{2021}{x}\right){dx}/{sinx} \\ $$

Answered by mnjuly1970 last updated on 15/Apr/21

𝛗_n =∫_0 ^( π) ((sin(nx))/(sin(x))) ........(∗) 𝛗_(n−2) =∫_0 ^( π) ((sin((n−2)x))/(sin(x)))dx 𝛗_n −𝛗_(n−2) =∫_0 ^( π) ((2cos(((nx+nx−2x)/2))sin(((2x)/2)))/(sin(x)))dx =∫_0 ^( π) ((cos((n−1)x).sin(x))/(sin(x)))dx =(1/(n−1))∫_0 ^( π) cos((n−1)x) dx=0 𝛗_n =𝛗_(n−2) ......✓ (∗)......:: 𝛗_0 =0 , 𝛗_1 =π ,𝛗_2 =0 (✓).. 𝛗_n = { (( 0 if ; n:=even)),(( π if ; n:= odd )) :} .......... ∴ 𝛗_(2021) = π ..........

$$\:\boldsymbol{\phi}_{{n}} =\int_{\mathrm{0}} ^{\:\pi} \frac{{sin}\left({nx}\right)}{{sin}\left({x}\right)}\:……..\left(\ast\right) \\ $$$$\:\boldsymbol{\phi}_{{n}−\mathrm{2}} =\int_{\mathrm{0}} ^{\:\pi} \frac{{sin}\left(\left({n}−\mathrm{2}\right){x}\right)}{{sin}\left({x}\right)}{dx} \\ $$$$\:\:\boldsymbol{\phi}_{{n}} −\boldsymbol{\phi}_{{n}−\mathrm{2}} =\int_{\mathrm{0}} ^{\:\pi} \frac{\mathrm{2}{cos}\left(\frac{{nx}+{nx}−\mathrm{2}{x}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{2}{x}}{\mathrm{2}}\right)}{{sin}\left({x}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\:\pi} \frac{{cos}\left(\left({n}−\mathrm{1}\right){x}\right).{sin}\left({x}\right)}{{sin}\left({x}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{n}−\mathrm{1}}\int_{\mathrm{0}} ^{\:\pi} {cos}\left(\left({n}−\mathrm{1}\right){x}\right)\:{dx}=\mathrm{0} \\ $$$$\:\:\boldsymbol{\phi}_{{n}} =\boldsymbol{\phi}_{{n}−\mathrm{2}} \:……\checkmark \\ $$$$\:\:\left(\ast\right)……::\:\:\boldsymbol{\phi}_{\mathrm{0}} =\mathrm{0}\:,\:\boldsymbol{\phi}_{\mathrm{1}} =\pi\:,\boldsymbol{\phi}_{\mathrm{2}} =\mathrm{0}\:\left(\checkmark\right).. \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}_{{n}} =\begin{cases}{\:\mathrm{0}\:\:\:\:\:\:{if}\:;\:\:{n}:={even}}\\{\:\pi\:\:\:\:\:\:{if}\:;\:\:{n}:=\:{odd}\:\:}\end{cases} \\ $$$$\:\:\:\:\:\:……….\:\:\:\therefore\:\:\:\:\boldsymbol{\phi}_{\mathrm{2021}} =\:\pi\:\:\:………. \\ $$$$\:\:\: \\ $$

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