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Question Number 122877 by bemath last updated on 20/Nov/20
∫ (sin^(−1) (x))^2 dx ?
∫(sin−1(x))2dx?
Commented by liberty last updated on 20/Nov/20
let u = sin^(−1) (x) ⇒x = sin u ⇒ dx = cos u du β(x)=∫ u^2 cos u du =u^2 sin u −2∫u sin u du = u^2 sin u−2(−ucos u +∫ cos u du) = u^2 sin u+2u cos u −2sin u + c = (u^2 −2)sin u +2u cos u + c = x[ (sin^(−1) (x))^2 −2 ]+2sin^(−1) (x) (√(1−x^2 )) + c
letu=sin−1(x)⇒x=sinu⇒dx=cosuduβ(x)=∫u2cosudu=u2sinu−2∫usinudu=u2sinu−2(−ucosu+∫cosudu)=u2sinu+2ucosu−2sinu+c=(u2−2)sinu+2ucosu+c=x[(sin−1(x))2−2]+2sin−1(x)1−x2+c
Answered by mathmax by abdo last updated on 21/Nov/20
A =∫ (arcsinx)^2 dx ⇒A =_(arcsinx=t) ∫ t^2 cost dt =t^2 sint−∫ 2t sint dt =t^2 sint −2 ∫ t sint dt =t^2 sint−2(−tcost +∫ cost dt) =t^2 sint+2t cost−2sint +C =x arcsin^2 x+2 arcsinx(√(1−t^2 ))−2x +C
A=∫(arcsinx)2dx⇒A=arcsinx=t∫t2costdt=t2sint−∫2tsintdt=t2sint−2∫tsintdt=t2sint−2(−tcost+∫costdt)=t2sint+2tcost−2sint+C=xarcsin2x+2arcsinx1−t2−2x+C

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