∫_(−π/2) ^(π/2) (x^2 +ln (((π+x)/(π−x))))cos x dx ?


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Question Number 121174 by benjo_mathlover last updated on 05/Nov/20

$\underset{- π / 2}{\int^{π / 2}} (x^{2} + \ln (\frac{π + x}{π - x})) \cos x dx ?$
	Answered by TANMAY PANACEA last updated on 05/Nov/20
	$I=∫_((−π)/2) ^(π/2) { (0−x)^2 +ln(((π+(0−x))/(π−(0−x))))}cos(0−x) dx I=∫_((−π)/2) ^(π/2) { x^2 +ln(((π−x)/(π+x)))}cosx dx 2I=∫_((−π)/2) ^(π/2) 2x^2 ×cosx+cosx{ln(((π+x)/(π−x)))+ln(((π−x)/(π+x)))}dx 2I=∫_((−π)/2) ^(π/2) 2x^2 ×cosx+cosx×ln1 dx I=∫_((−π)/2) ^(π/2) x^2 cosx dx J=∫x^2 e^(ix) dx =x^2 (e^(ix) /i)−∫2x×(e^(ix) /i)dx =((x^2 e^(ix) ×i)/(−1))+2i∫xe^(ix) dx =−ix^2 (cosx+isinx)+2i[x×(e^(ix) /i)−∫(e^(ix) /i)dx] =−ix^2 (cosx+isinx)+2x(cosx+isinx)−2×(e^(ix) /i) =(−ix^2 cosx)+x^2 sinx+2xcosx+i×2xcosx+2i(cosx+isinx) =real part +im part real psrt=(x^2 sinx+2xcosx−2sinx) I=∣x^2 sinx+2xcosx−2sinx∣_((−π)/2) ^(π/2) =(π^2 /4)×1−(π^2 /4)×(−1)+2×(π/2)×0−2×((−π)/2)×0−2(1+1) =(π^2 /2)−4$
	$I = \int_{\frac{- π}{2}}^{\frac{π}{2}} {{(0 - x)}^{2} + l n (\frac{π + (0 - x)}{π - (0 - x)})} c o s (0 - x) d x$ $I = \int_{\frac{- π}{2}}^{\frac{π}{2}} {x^{2} + l n (\frac{π - x}{π + x})} c o s x d x$ $2 I = \int_{\frac{- π}{2}}^{\frac{π}{2}} 2 x^{2} \times c o s x + c o s x {l n (\frac{π + x}{π - x}) + l n (\frac{π - x}{π + x})} d x$ $2 I = \int_{\frac{- π}{2}}^{\frac{π}{2}} 2 x^{2} \times c o s x + c o s x \times l n 1 d x$ $I = \int_{\frac{- π}{2}}^{\frac{π}{2}} x^{2} c o s x d x$ $J = \int x^{2} e^{i x} d x$ $= x^{2} \frac{e^{i x}}{i} - \int 2 x \times \frac{e^{i x}}{i} d x$ $= \frac{x^{2} e^{i x} \times i}{- 1} + 2 i \int {x e}^{i x} d x$ $= - {i x}^{2} (c o s x + i s i n x) + 2 i [x \times \frac{e^{i x}}{i} - \int \frac{e^{i x}}{i} d x]$ $= - {i x}^{2} (c o s x + i s i n x) + 2 x (c o s x + i s i n x) - 2 \times \frac{e^{i x}}{i}$ $= (- {i x}^{2} c o s x) + x^{2} s i n x + 2 x c o s x + i \times 2 x c o s x + 2 i (c o s x + i s i n x)$ $= r e a l p a r t + i m p a r t$ $r e a l p s r t = (x^{2} s i n x + 2 x c o s x - 2 s i n x)$ $I =∣ x^{2} s i n x + 2 x c o s x - 2 s i n x ∣_{\frac{- π}{2}}^{\frac{π}{2}}$ $= \frac{π^{2}}{4} \times 1 - \frac{π^{2}}{4} \times (- 1) + 2 \times \frac{π}{2} \times 0 - 2 \times \frac{- π}{2} \times 0 - 2 (1 + 1)$ $= \frac{π^{2}}{2} - 4$
	Commented by TANMAY PANACEA last updated on 05/Nov/20

	$y e s y e s i c o u l d n o t s e e . . .$
	Commented by benjo_mathlover last updated on 06/Nov/20

	$yess . . .$
	Answered by Ar Brandon last updated on 05/Nov/20
	$I=∫_(−(π/2)) ^(π/2) {x^2 +ln(((π+x)/(π−x)))}cosxdx I=∫_(−(π/2)) ^(π/2) {x^2 +ln(((π−x)/(π+x)))}cosxdx I+I=∫_(−(π/2)) ^(π/2) {2x^2 +ln(((π+x)/(π−x)))+ln(((π−x)/(π+x)))}cosxdx ⇒2I=∫_(−(π/2)) ^(π/2) 2x^2 cosxdx=2[x^2 ∫cosxdx−∫{(dx^2 /dx)∙∫cosxdx}]_(−(π/2)) ^(π/2) ⇒I=[x^2 sinx−2∫xsinxdx]_(−(π/2)) ^(π/2) =[x^2 sinx+2{xcosx−∫cosxdx}]_(−(π/2)) ^(π/2) =[x^2 sinx+2xcosx−2sinx]_(−(π/2)) ^(π/2) ={(π^2 /4)−2}+{(π^2 /4)−2} =(π^2 /2)−4$
	$I = \int_{- \frac{π}{2}}^{\frac{π}{2}} {x^{2} + \ln (\frac{π + x}{π - x})} cosxdx$ $I = \int_{- \frac{π}{2}}^{\frac{π}{2}} {x^{2} + \ln (\frac{π - x}{π + x})} cosxdx$ $I + I = \int_{- \frac{π}{2}}^{\frac{π}{2}} {{2 x}^{2} + \ln (\frac{π + x}{π - x}) + \ln (\frac{π - x}{π + x})} cosxdx$ $\Rightarrow 2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} {2 x}^{2} cosxdx = 2 {[x^{2} \int cosxdx - \int {\frac{{dx}^{2}}{dx} \cdot \int cosxdx}]}_{- \frac{π}{2}}^{\frac{π}{2}}$ $\Rightarrow I = {[x^{2} sinx - 2 \int xsinxdx]}_{- \frac{π}{2}}^{\frac{π}{2}}$ $= {[x^{2} sinx + 2 {xcosx - \int cosxdx}]}_{- \frac{π}{2}}^{\frac{π}{2}}$ $= {[x^{2} sinx + 2 xcosx - 2 sinx]}_{- \frac{π}{2}}^{\frac{π}{2}} = {\frac{π^{2}}{4} - 2} + {\frac{π^{2}}{4} - 2}$ $= \frac{π^{2}}{2} - 4$
	Commented by Ar Brandon last updated on 05/Nov/20

	Commented by Ar Brandon last updated on 05/Nov/20

	$\int_{a}^{b} f (x) dx = \int_{a}^{b} f (a + b - x) dx$
	Answered by Dwaipayan Shikari last updated on 05/Nov/20

	$I n t e g r a l i s s y m m e t r i c$ $\int_{- \frac{π}{2}}^{\frac{π}{2}} l o g (\frac{π + x}{π - x}) c o s x = 0$ $\int_{- \frac{π}{2}}^{\frac{π}{2}} x^{2} c o s x d x$ $= {[x^{2} s i n x]}_{- \frac{π}{2}}^{\frac{π}{2}} - \int_{- \frac{π}{2}}^{\frac{π}{2}} 2 x s i n x d x$ $= \frac{π^{2}}{2} + 2 {[x c o s x]}_{- \frac{π}{2}}^{\frac{π}{2}} - 2 \int_{- \frac{π}{2}}^{\frac{π}{2}} c o s x$ $= \frac{π^{2}}{2} - 2.2 = \frac{π^{2}}{2} - 4$
	Answered by Bird last updated on 05/Nov/20
	$∫_(−(π/2)) ^(π/2) (x^2 +ln(((π+x)/(π−x))))cosx dx =∫_(−(π/2)) ^(π/2) x^2 cosx dx+∫_(−(π/2)) ^(π/2) ln(((π+x)/(π−x)))cosxdx(→=0 odd function under integral) =∫_(−(π/2)) ^(π/2) x^2 cosx dx=[x^2 sinx]_(−(π/2)) ^(π/2) −∫_(−(π/2)) ^(π/2) 2x sinx dx =(π^2 /2)−2 ∫_(−(π/2)) ^(π/2) xsinx dx we have ∫_(−(π/2)) ^(π/2) xsinx dx =2∫_0 ^(π/2) xsinx dx =2{[−xcosx]_0 ^(π/2) +∫_0 ^(π/2) cosxdx} =2 [sinx]_0 ^(π/2) =2 ⇒ I =(π^2 /2)−4$
	$\int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{2} + l n (\frac{π + x}{π - x})) c o s x d x$ $= \int_{- \frac{π}{2}}^{\frac{π}{2}} x^{2} c o s x d x + \int_{- \frac{π}{2}}^{\frac{π}{2}} l n (\frac{π + x}{π - x}) c o s x d x (\to= 0 o d d f u n c t i o n u n d e r i n t e g r a l)$ $= \int_{- \frac{π}{2}}^{\frac{π}{2}} x^{2} c o s x d x = {[x^{2} s i n x]}_{- \frac{π}{2}}^{\frac{π}{2}}$ $- \int_{- \frac{π}{2}}^{\frac{π}{2}} 2 x s i n x d x$ $= \frac{π^{2}}{2} - 2 \int_{- \frac{π}{2}}^{\frac{π}{2}} x s i n x d x w e h a v e$ $\int_{- \frac{π}{2}}^{\frac{π}{2}} x s i n x d x = 2 \int_{0}^{\frac{π}{2}} x s i n x d x$ $= 2 {{[- x c o s x]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} c o s x d x}$ $= 2 {[s i n x]}_{0}^{\frac{π}{2}} = 2 \Rightarrow$ $I = \frac{π^{2}}{2} - 4$
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