I-0-pi-2-x-2-1-sin-2-x-dx-

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 \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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