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Question Number 174071 by alcohol last updated on 24/Jul/22
Answered by aleks041103 last updated on 24/Jul/22
.1.F(x)=∫0xdt1−t2=∫0xF′(t)dt⇒F′(x)=11−x22.F(x)=∫0xdt1−t2=arcsin(t)∣0xF(x)=arcsin(x)⇒G(x)=−arcsin(x)F(sin(x))=arcsin(sin(x))=x3.G(cos(x))=−arcsin(cos(x))−arcsin(t)=arccos(t)−π/2⇒G(cos(x))=x−π24.FisaninversetothesinGisnotaninverseforthecos
Answered by CElcedricjunior last updated on 25/Jul/22
resolution1)F′(x)=11−x2−12)F(sin(x))=∫0sinxdt1−t2posonst=sinadt=cosadaqd:{t=sinxt=0=>{a=xt=0F(sinx)=∫0xcosada−sinx2=∫0xda=xF(sinx)=x3)G(cosx)=−x+π/2
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