Question Number 121174 by benjo_mathlover last updated on 05/Nov/20
$$\underset{−\pi/\mathrm{2}} {\overset{\pi/\mathrm{2}} {\int}}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{ln}\:\left(\frac{\pi+\mathrm{x}}{\pi−\mathrm{x}}\right)\right)\mathrm{cos}\:\mathrm{x}\:\mathrm{dx}\:? \\ $$
Answered by TANMAY PANACEA last updated on 05/Nov/20
![I=∫_((−π)/2) ^(π/2) { (0−x)^2 +ln(((π+(0−x))/(π−(0−x))))}cos(0−x) dx I=∫_((−π)/2) ^(π/2) { x^2 +ln(((π−x)/(π+x)))}cosx dx 2I=∫_((−π)/2) ^(π/2) 2x^2 ×cosx+cosx{ln(((π+x)/(π−x)))+ln(((π−x)/(π+x)))}dx 2I=∫_((−π)/2) ^(π/2) 2x^2 ×cosx+cosx×ln1 dx I=∫_((−π)/2) ^(π/2) x^2 cosx dx J=∫x^2 e^(ix) dx =x^2 (e^(ix) /i)−∫2x×(e^(ix) /i)dx =((x^2 e^(ix) ×i)/(−1))+2i∫xe^(ix) dx =−ix^2 (cosx+isinx)+2i[x×(e^(ix) /i)−∫(e^(ix) /i)dx] =−ix^2 (cosx+isinx)+2x(cosx+isinx)−2×(e^(ix) /i) =(−ix^2 cosx)+x^2 sinx+2xcosx+i×2xcosx+2i(cosx+isinx) =real part +im part real psrt=(x^2 sinx+2xcosx−2sinx) I=∣x^2 sinx+2xcosx−2sinx∣_((−π)/2) ^(π/2) =(π^2 /4)×1−(π^2 /4)×(−1)+2×(π/2)×0−2×((−π)/2)×0−2(1+1) =(π^2 /2)−4](https://www.tinkutara.com/question/Q121176.png)
$${I}=\int_{\frac{−\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left\{\:\left(\mathrm{0}−{x}\right)^{\mathrm{2}} +{ln}\left(\frac{\pi+\left(\mathrm{0}−{x}\right)}{\pi−\left(\mathrm{0}−{x}\right)}\right)\right\}{cos}\left(\mathrm{0}−{x}\right)\:{dx} \\ $$$${I}=\int_{\frac{−\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left\{\:{x}^{\mathrm{2}} +{ln}\left(\frac{\pi−{x}}{\pi+{x}}\right)\right\}{cosx}\:{dx} \\ $$$$\mathrm{2}{I}=\int_{\frac{−\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{x}^{\mathrm{2}} ×{cosx}+{cosx}\left\{{ln}\left(\frac{\pi+{x}}{\pi−{x}}\right)+{ln}\left(\frac{\pi−{x}}{\pi+{x}}\right)\right\}{dx} \\ $$$$\mathrm{2}{I}=\int_{\frac{−\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{x}^{\mathrm{2}} ×{cosx}+{cosx}×{ln}\mathrm{1}\:\:{dx} \\ $$$${I}=\int_{\frac{−\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{2}} {cosx}\:{dx} \\ $$$${J}=\int{x}^{\mathrm{2}} {e}^{{ix}} {dx} \\ $$$$={x}^{\mathrm{2}} \frac{{e}^{{ix}} }{{i}}−\int\mathrm{2}{x}×\frac{{e}^{{ix}} }{{i}}{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} {e}^{{ix}} ×{i}}{−\mathrm{1}}+\mathrm{2}{i}\int{xe}^{{ix}} {dx} \\ $$$$=−{ix}^{\mathrm{2}} \left({cosx}+{isinx}\right)+\mathrm{2}{i}\left[{x}×\frac{{e}^{{ix}} }{{i}}−\int\frac{{e}^{{ix}} }{{i}}{dx}\right] \\ $$$$=−{ix}^{\mathrm{2}} \left({cosx}+{isinx}\right)+\mathrm{2}{x}\left({cosx}+{isinx}\right)−\mathrm{2}×\frac{{e}^{{ix}} }{{i}} \\ $$$$=\left(−{ix}^{\mathrm{2}} {cosx}\right)+{x}^{\mathrm{2}} {sinx}+\mathrm{2}{xcosx}+{i}×\mathrm{2}{xcosx}+\mathrm{2}{i}\left({cosx}+{isinx}\right) \\ $$$$={real}\:{part}\:+{im}\:{part} \\ $$$$ \\ $$$${real}\:{psrt}=\left({x}^{\mathrm{2}} {sinx}+\mathrm{2}{xcosx}−\mathrm{2}{sinx}\right) \\ $$$${I}=\mid{x}^{\mathrm{2}} {sinx}+\mathrm{2}{xcosx}−\mathrm{2}{sinx}\mid_{\frac{−\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}×\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}×\left(−\mathrm{1}\right)+\mathrm{2}×\frac{\pi}{\mathrm{2}}×\mathrm{0}−\mathrm{2}×\frac{−\pi}{\mathrm{2}}×\mathrm{0}−\mathrm{2}\left(\mathrm{1}+\mathrm{1}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\mathrm{4} \\ $$
Commented by TANMAY PANACEA last updated on 05/Nov/20
$${yes}\:{yes}\:{i}\:{could}\:{not}\:{see}… \\ $$
Commented by benjo_mathlover last updated on 06/Nov/20
$$\mathrm{yess}… \\ $$
Answered by Ar Brandon last updated on 05/Nov/20
![I=∫_(−(π/2)) ^(π/2) {x^2 +ln(((π+x)/(π−x)))}cosxdx I=∫_(−(π/2)) ^(π/2) {x^2 +ln(((π−x)/(π+x)))}cosxdx I+I=∫_(−(π/2)) ^(π/2) {2x^2 +ln(((π+x)/(π−x)))+ln(((π−x)/(π+x)))}cosxdx ⇒2I=∫_(−(π/2)) ^(π/2) 2x^2 cosxdx=2[x^2 ∫cosxdx−∫{(dx^2 /dx)∙∫cosxdx}]_(−(π/2)) ^(π/2) ⇒I=[x^2 sinx−2∫xsinxdx]_(−(π/2)) ^(π/2) =[x^2 sinx+2{xcosx−∫cosxdx}]_(−(π/2)) ^(π/2) =[x^2 sinx+2xcosx−2sinx]_(−(π/2)) ^(π/2) ={(π^2 /4)−2}+{(π^2 /4)−2} =(π^2 /2)−4](https://www.tinkutara.com/question/Q121177.png)
$$\mathcal{I}=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left\{\mathrm{x}^{\mathrm{2}} +\mathrm{ln}\left(\frac{\pi+\mathrm{x}}{\pi−\mathrm{x}}\right)\right\}\mathrm{cosxdx} \\ $$$$\mathcal{I}=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left\{\mathrm{x}^{\mathrm{2}} +\mathrm{ln}\left(\frac{\pi−\mathrm{x}}{\pi+\mathrm{x}}\right)\right\}\mathrm{cosxdx} \\ $$$$\mathcal{I}+\mathcal{I}=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left\{\mathrm{2x}^{\mathrm{2}} +\mathrm{ln}\left(\frac{\pi+\mathrm{x}}{\pi−\mathrm{x}}\right)+\mathrm{ln}\left(\frac{\pi−\mathrm{x}}{\pi+\mathrm{x}}\right)\right\}\mathrm{cosxdx} \\ $$$$\Rightarrow\mathrm{2}\mathcal{I}=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2x}^{\mathrm{2}} \mathrm{cosxdx}=\mathrm{2}\left[\mathrm{x}^{\mathrm{2}} \int\mathrm{cosxdx}−\int\left\{\frac{\mathrm{dx}^{\mathrm{2}} }{\mathrm{dx}}\centerdot\int\mathrm{cosxdx}\right\}\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\Rightarrow\mathcal{I}=\left[\mathrm{x}^{\mathrm{2}} \mathrm{sinx}−\mathrm{2}\int\mathrm{xsinxdx}\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:=\left[\mathrm{x}^{\mathrm{2}} \mathrm{sinx}+\mathrm{2}\left\{\mathrm{xcosx}−\int\mathrm{cosxdx}\right\}\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:=\left[\mathrm{x}^{\mathrm{2}} \mathrm{sinx}+\mathrm{2xcosx}−\mathrm{2sinx}\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} =\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\mathrm{2}\right\}+\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\mathrm{2}\right\} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\mathrm{4} \\ $$
Commented by Ar Brandon last updated on 05/Nov/20
Commented by Ar Brandon last updated on 05/Nov/20
$$\int_{\mathrm{a}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{a}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{a}+\mathrm{b}−\mathrm{x}\right)\mathrm{dx} \\ $$
Answered by Dwaipayan Shikari last updated on 05/Nov/20
![Integral is symmetric ∫_(−(π/2)) ^(π/2) log(((π+x)/(π−x)))cosx=0 ∫_(−(π/2)) ^(π/2) x^2 cosxdx =[x^2 sinx]_(−(π/2)) ^(π/2) −∫_(−(π/2)) ^(π/2) 2xsinxdx =(π^2 /2)+2[xcosx]_(−(π/2)) ^(π/2) −2∫_(−(π/2)) ^(π/2) cosx =(π^2 /2)−2.2 =(π^2 /2)−4](https://www.tinkutara.com/question/Q121178.png)
$${Integral}\:{is}\:{symmetric}\: \\ $$$$\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {log}\left(\frac{\pi+{x}}{\pi−{x}}\right){cosx}=\mathrm{0} \\ $$$$\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{2}} {cosxdx} \\ $$$$=\left[{x}^{\mathrm{2}} {sinx}\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} −\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{xsinxdx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}\left[{xcosx}\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} −\mathrm{2}\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {cosx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}.\mathrm{2}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\mathrm{4} \\ $$
Answered by Bird last updated on 05/Nov/20
![∫_(−(π/2)) ^(π/2) (x^2 +ln(((π+x)/(π−x))))cosx dx =∫_(−(π/2)) ^(π/2) x^2 cosx dx+∫_(−(π/2)) ^(π/2) ln(((π+x)/(π−x)))cosxdx(→=0 odd function under integral) =∫_(−(π/2)) ^(π/2) x^2 cosx dx=[x^2 sinx]_(−(π/2)) ^(π/2) −∫_(−(π/2)) ^(π/2) 2x sinx dx =(π^2 /2)−2 ∫_(−(π/2)) ^(π/2) xsinx dx we have ∫_(−(π/2)) ^(π/2) xsinx dx =2∫_0 ^(π/2) xsinx dx =2{[−xcosx]_0 ^(π/2) +∫_0 ^(π/2) cosxdx} =2 [sinx]_0 ^(π/2) =2 ⇒ I =(π^2 /2)−4](https://www.tinkutara.com/question/Q121192.png)
$$\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left({x}^{\mathrm{2}} \:+{ln}\left(\frac{\pi+{x}}{\pi−{x}}\right)\right){cosx}\:{dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{2}} {cosx}\:{dx}+\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\pi+{x}}{\pi−{x}}\right){cosxdx}\left(\rightarrow=\mathrm{0}\:\:{odd}\:{function}\:{under}\:{integral}\right) \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{x}^{\mathrm{2}} {cosx}\:{dx}=\left[{x}^{\mathrm{2}} {sinx}\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$−\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{x}\:{sinx}\:{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{xsinx}\:{dx}\:\:{we}\:{have} \\ $$$$\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{xsinx}\:{dx}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {xsinx}\:{dx} \\ $$$$=\mathrm{2}\left\{\left[−{xcosx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cosxdx}\right\} \\ $$$$=\mathrm{2}\:\left[{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\mathrm{2}\:\Rightarrow \\ $$$${I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\mathrm{4} \\ $$$$ \\ $$