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0-1-x-2n-arcsinx-dx-




Question Number 156339 by lapache last updated on 10/Oct/21
∫_0 ^1 (x^(2n) /(arcsinx))dx=...???
01x2narcsinxdx=???
Answered by ArielVyny last updated on 10/Oct/21
t=arcsinx→sint=x  ∫_0 ^(π/2) ((sin^(2n) t)/t)costdt=∫_0 ^(π/2) ((sin^(2n) t)/t)Σ_(n≥0) (((−1)^n t^(2n) )/((2n)!))  Σ_(n≥0) (((−1)^n )/((2n)!))∫_0 ^(π/2) t^(2n) sin^(2n) tdt
t=arcsinxsint=x0π2sin2nttcostdt=0π2sin2nttn0(1)nt2n(2n)!n0(1)n(2n)!0π2t2nsin2ntdt

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