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Question Number 217755 by Tawa11 last updated on 20/Mar/25
∫ ((cos(sin^(− 1) x) + cos^(− 1) (sin x))/(ln(ln(ln(1 + (√(x + (√x)))))) dx
cos(sin1x)+cos1(sinx)ln(ln(ln(1+x+x)dx
Commented by mr W last updated on 20/Mar/25
you can even make it more nice  looking  ∫ ((cos(sin^(− 1) x+e^(tan x) ) + cos^(− 1) (sin x+tanh x))/(ln(ln(ln(1 + (√(x + (√x)))))) dx  there are a lot of other mathematic  functions and symbols which can  still be put into the formula.
youcanevenmakeitmorenicelookingcos(sin1x+etanx)+cos1(sinx+tanhx)ln(ln(ln(1+x+x)dxtherearealotofothermathematicfunctionsandsymbolswhichcanstillbeputintotheformula.
Commented by Tawa11 last updated on 20/Mar/25
Hahahahaha!!!
Hahahahaha!!!
Answered by SdC355 last updated on 20/Mar/25
Hint.  cos(asin(z))=(√(1−z^2 ))  sin^2 (θ)+cos^2 (θ)=1  sin(θ)=±(√(1−cos^2 (θ)))   and θ=acos(z)  sin(acos(z))=±(√(1−cos^2 (acos(z))))=(√(1−z^2 ))  ∵ f∗f^(−1) =z  {∗ is function composition operator}  cos^(−1) (sin(z))=(π/2)−z  i think no one can solve ∫  dz  (((√(1−z^2 ))+(π/2)−z)/(ln(ln(ln(1+(√(z+(√z))))))))
Hint.cos(asin(z))=1z2sin2(θ)+cos2(θ)=1sin(θ)=±1cos2(θ)andθ=acos(z)sin(acos(z))=±1cos2(acos(z))=1z2ff1=z{isfunctioncompositionoperator}cos1(sin(z))=π2zithinknoonecansolvedz1z2+π2zln(ln(ln(1+z+z)))
Commented by Tawa11 last updated on 20/Mar/25
That is serious sir.
Thatisserioussir.
Commented by SdC355 last updated on 20/Mar/25
Because  ∫  (((√(1−z^2 ))+(π/2)−z)/(′′ln(ln(ln(1+(√(z+(√z))))))′′))
Because1z2+π2zln(ln(ln(1+z+z)))
Answered by Marzuk last updated on 20/Mar/25
Nah boy.We might put some function acrylic  here it is.  ∫ (((λx.Γ(x)+β(x,x+1)+Θ(x)+α(x).ℜ(e^(i cos^(−1) x) )+ (∂/∂x) [ Γ(x)+β(x,x+1)+Θ(x))/(ln(ln(ln(1+2(√x))))+cosh^(−1) (1+x^2 ) ∫_0 ^( x) (dt/(1+t^2 )) det (((x      cos(x))),((sin(x)    x^2 )) ) + ψ(x)+Θ(x)+α(x))) dx  Now the painting is perfect
Nahboy.Wemightputsomefunctionacrylichereitis.(λx.Γ(x)+β(x,x+1)+Θ(x)+α(x).(eicos1x)+x[Γ(x)+β(x,x+1)+Θ(x)ln(ln(ln(1+2x)))+cosh1(1+x2)0xdt1+t2det(xcos(x)sin(x)x2)+ψ(x)+Θ(x)+α(x)dxNowthepaintingisperfect

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