I = ∫_0 ^(π/2) (x^2 /(1+sin^2 (x)))dx


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Question Number 203615 by muneer0o0 last updated on 23/Jan/24

$I = \int_{0}^{π / 2} \frac{x^{2}}{1 + \sin^{2} (x)} d x$
	Answered by witcher3 last updated on 25/Jan/24
	$I=Re∫_0 ^(π/2) (x^2 /(1+isin(x)))dx=Re(B) B=∫_0 ^(π/2) ((2x^2 )/(2+e^(ix) −e^(−ix) ))=∫_0 ^(π/2) ((2x^2 e^(ix) )/((e^(2ix) +2e^(ix) −1)))dx=∫_0 ^(π/2) ((2x^2 e^(ix) )/((e^(ix) +1+(√2))(e^(ix) +1−(√2))))dx =2∫_0 ^(π/2) x^2 [.(((−1−(√2))/(−2(√2)))).(1/((e^(ix) +1+(√2))))+((−1+(√2))/(2(√2)(e^(ix) +1−(√2))))]dx =(1/( (√2)))∫_0 ^(π/2) (x^2 /((e^(ix) /(1+(√2)))+1))dx+(((√2)−1)/( (√2))).∫_0 ^(π/2) ((x^2 e^(−ix) )/(1+(1−(√2))e^(−ix) ))dx =(1/( (√2)))∫_0 ^(π/2) Σ(((−1)^k )/((1+(√2))^k )).x^2 e^(ikx) dx+(1/( (√2)))∫_0 ^(π/2) ((√2)−1)^(k+1) e^(−i(k+1)x) x^2 dx =(1/( (√2)))∫_0 ^(π/2) Σ_(k≥0) x^2 (1−(√2))^k x^2 e^(ikx) dx+(1/( (√2)))∫_0 ^(π/2) ((√2)−1)^(k+1) e^(−i(k+1)x) x^2 dx we want Re (B);∫x^2 e^(ikx) dxor ∫x^2 e^(−ikx) dx didnt change realart⇒ B=(1/( (√2)))((π^3 /(24)))+(1/( (√2)))Σ_(k≥0) (1−(√2))^(k+1) x^2 e^(−i(k+1)x) +((√2)−1)^(k+1) e =(1/( (√2))).(π^3 /(24))+(2/( (√2))) Σ_(k≥0) ((√2)−1)^(2k+2) e^(−i(2k+2)x) x^2 dx =(1/( (√2))).(π^3 /(24))+(√2).Σ((√2)−1)^(2k+2) ∫_0 ^π e^(i(k+1)y) .(y^2 /8)dy =(π^3 /( 24(√2)))+(1/(2(√2)))Σ((√2)−1)^(2k+2) {.[((y^2 e^(i(k+1)y) )/(i(k+1)))]_0 ^π −(2/(i(k+1)))∫_0 ^π ye^(i(k+1)y) dy =(π^3 /( (√(242))))+(1/( (√2)))Σ((√2)−1)^(2k+2) .[((ye^(i(k+1)y) )/((k+1)^2 ))]_0 ^π =(π^3 /( 24(√2)))+(1/( (√2)))Σ(((3−2(√2))^(k+1) π(−1)^(k+1) )/((k+1)^2 )) =(π^3 /( 24(√2)))+(π/( (√2)))Li_2 (2(√2)−3);Li_s (z)=Σ_(k≥0) (z^(k+1) /((k+1)^s )),∀∣z∣<1 ∫_0 ^(π/2) (x^2 /(1+sin^2 (x)))=(π^3 /(24(√2)))+(π/( (√2)))Li_2 (2(√2)−3)$
	$I = Re \int_{0}^{\frac{π}{2}} \frac{x^{2}}{1 + isin (x)} dx = Re (B)$ $B = \int_{0}^{\frac{π}{2}} \frac{{2 x}^{2}}{2 + e^{ix} - e^{- ix}} = \int_{0}^{\frac{π}{2}} \frac{{2 x}^{2} e^{ix}}{(e^{2 ix} + {2 e}^{ix} - 1)} dx = \int_{0}^{\frac{π}{2}} \frac{{2 x}^{2} e^{ix}}{(e^{ix} + 1 + \sqrt{2}) (e^{ix} + 1 - \sqrt{2})} dx$ $= 2 \int_{0}^{\frac{π}{2}} x^{2} [. (\frac{- 1 - \sqrt{2}}{- 2 \sqrt{2}}) . \frac{1}{(e^{ix} + 1 + \sqrt{2})} + \frac{- 1 + \sqrt{2}}{2 \sqrt{2} (e^{ix} + 1 - \sqrt{2})}] dx$ $= \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{x^{2}}{\frac{e^{ix}}{1 + \sqrt{2}} + 1} dx + \frac{\sqrt{2} - 1}{\sqrt{2}} . \int_{0}^{\frac{π}{2}} \frac{x^{2} e^{- ix}}{1 + (1 - \sqrt{2}) e^{- ix}} dx$ $= \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} Σ \frac{{(- 1)}^{k}}{{(1 + \sqrt{2})}^{k}} . x^{2} e^{ikx} dx + \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} {(\sqrt{2} - 1)}^{k + 1} e^{- i (k + 1) x} x^{2} dx$ $= \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \sum_{k ⩾ 0} x^{2} {(1 - \sqrt{2})}^{k} x^{2} e^{ikx} dx + \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} {(\sqrt{2} - 1)}^{k + 1} e^{- i (k + 1) x} x^{2} dx$ $we want Re (B); \int x^{2} e^{ikx} dxor \int x^{2} e^{- ikx} dx didnt change realart \Rightarrow$ $B = \frac{1}{\sqrt{2}} (\frac{π^{3}}{24}) + \frac{1}{\sqrt{2}} \sum_{k ⩾ 0} {(1 - \sqrt{2})}^{k + 1} x^{2} e^{- i (k + 1) x} + {(\sqrt{2} - 1)}^{k + 1} e$ $= \frac{1}{\sqrt{2}} . \frac{π^{3}}{24} + \frac{2}{\sqrt{2}} \sum_{k ⩾ 0} {(\sqrt{2} - 1)}^{2 k + 2} e^{- i (2 k + 2) x} x^{2} dx$ $= \frac{1}{\sqrt{2}} . \frac{π^{3}}{24} + \sqrt{2} . Σ {(\sqrt{2} - 1)}^{2 k + 2} \int_{0}^{π} e^{i (k + 1) y} . \frac{y^{2}}{8} dy$ $= \frac{π^{3}}{24 \sqrt{2}} + \frac{1}{2 \sqrt{2}} Σ {(\sqrt{2} - 1)}^{2 k + 2} {. {[\frac{y^{2} e^{i (k + 1) y}}{i (k + 1)}]}_{0}^{π} - \frac{2}{i (k + 1)} \int_{0}^{π} {ye}^{i (k + 1) y} dy$ $= \frac{π^{3}}{\sqrt{242}} + \frac{1}{\sqrt{2}} Σ {(\sqrt{2} - 1)}^{2 k + 2} . {[\frac{{ye}^{i (k + 1) y}}{{(k + 1)}^{2}}]}_{0}^{π}$ $= \frac{π^{3}}{24 \sqrt{2}} + \frac{1}{\sqrt{2}} Σ \frac{{(3 - 2 \sqrt{2})}^{k + 1} π {(- 1)}^{k + 1}}{{(k + 1)}^{2}}$ $= \frac{π^{3}}{24 \sqrt{2}} + \frac{π}{\sqrt{2}} {Li}_{2} (2 \sqrt{2} - 3); {Li}_{s} (z) = \sum_{k ⩾ 0} \frac{z^{k + 1}}{{(k + 1)}^{s}}, \forall ∣ z ∣< 1$ $\int_{0}^{\frac{π}{2}} \frac{x^{2}}{1 + \sin^{2} (x)} = \frac{π^{3}}{24 \sqrt{2}} + \frac{π}{\sqrt{2}} {Li}_{2} (2 \sqrt{2} - 3)$
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