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Question Number 137698 by JulioCesar last updated on 05/Apr/21

Commented by peter frank last updated on 06/Apr/21

you forgot to set a limt

$${you}\:{forgot}\:{to}\:{set}\:{a}\:{limt} \\ $$

Answered by peter frank last updated on 06/Apr/21

∫_0 ^π ((xsin x)/(1+cos^2 x))dx

$$\int_{\mathrm{0}} ^{\pi} \frac{{x}\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {x}}{dx} \\ $$

Commented by SEKRET last updated on 07/Apr/21

∫_a ^b f(x) dx=∫_a ^b f(a+b+x) dx= I ∫_0 ^𝛑 ((x∙sin(x))/(1+cos^2 (x))) dx=∫_0 ^𝛑 (((𝛑−x)∙sin(𝛑−x))/(1+cos^2 (𝛑−x))) dx ∫_0 ^𝛑 ((x∙sin(x))/(1+cos^2 (x)))dx=∫_0 ^𝛑 ((𝛑∙sin(x))/(1+cos^2 (x))) dx−∫_0 ^𝛑 ((x∙sin(x))/(1+cos^2 (x)))dx I=𝛑∙∫_0 ^𝛑 ((sin(x))/(1+cos^2 (x))) dx − I ((2I)/𝛑) = ∫_0 ^𝛑 (( − d(cos(x)))/(1+cos^2 (x))) dx ((2I)/𝛑) = −arctg(cos(x)) /_0 ^𝛑 = (𝛑/4)−(((−𝛑)/4))=(𝛑/2) ((2I)/𝛑) = (𝛑/2) I = (𝛑^2 /4) ∫_0 ^𝛑 ((x∙sin(x))/(1+cos^2 (x)))dx= ((−𝛑^2 )/4) ABDULAZIZ ABDUVALIYEV

$$\int_{\boldsymbol{\mathrm{a}}} ^{\boldsymbol{\mathrm{b}}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{dx}}=\int_{\boldsymbol{\mathrm{a}}} ^{\boldsymbol{\mathrm{b}}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{dx}}=\:\boldsymbol{\mathrm{I}} \\ $$$$\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \frac{\boldsymbol{\mathrm{x}}\centerdot\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}\:\boldsymbol{\mathrm{dx}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \frac{\left(\boldsymbol{\pi}−\boldsymbol{\mathrm{x}}\right)\centerdot\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\pi}−\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\pi}−\boldsymbol{\mathrm{x}}\right)}\:\boldsymbol{\mathrm{dx}} \\ $$$$\:\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \frac{\boldsymbol{\mathrm{x}}\centerdot\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}\boldsymbol{\mathrm{dx}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \frac{\boldsymbol{\pi}\centerdot\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}\:\boldsymbol{\mathrm{dx}}−\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \frac{\boldsymbol{\mathrm{x}}\centerdot\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}\boldsymbol{\mathrm{dx}} \\ $$$$\:\boldsymbol{\mathrm{I}}=\boldsymbol{\pi}\centerdot\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}\:\boldsymbol{\mathrm{dx}}\:−\:\boldsymbol{\mathrm{I}} \\ $$$$\:\frac{\mathrm{2}\boldsymbol{\mathrm{I}}}{\boldsymbol{\pi}}\:=\:\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \frac{\:−\:\boldsymbol{\mathrm{d}}\left(\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)\right)}{\mathrm{1}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}\:\boldsymbol{\mathrm{dx}} \\ $$$$\:\frac{\mathrm{2}\boldsymbol{\mathrm{I}}}{\boldsymbol{\pi}}\:\:\:=\:\:−\boldsymbol{\mathrm{arctg}}\left(\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)\right)\:/_{\mathrm{0}} ^{\boldsymbol{\pi}} \:=\:\frac{\boldsymbol{\pi}}{\mathrm{4}}−\left(\frac{−\boldsymbol{\pi}}{\mathrm{4}}\right)=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\:\:\:\:\frac{\mathrm{2}\boldsymbol{\mathrm{I}}}{\boldsymbol{\pi}}\:\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{I}}\:=\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\int_{\mathrm{0}} ^{\boldsymbol{\pi}} \:\frac{\boldsymbol{\mathrm{x}}\centerdot\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{x}}\right)}\boldsymbol{\mathrm{dx}}=\:\frac{−\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{ABDULAZIZ}}\:\:\:\boldsymbol{\mathrm{ABDUVALIYEV}} \\ $$

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