What is the integral of x^3 over square rood of 1+x^2 dx by using the substitution u=sinh (x)?

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Answer 1 · 2018-03-08T19:15:07

The answer is #=1/3(x^2-2)sqrt(x^2+1)+C#

We need

#cosh^2x-sinh^2x=1#

#arcsinhx=ln(x+sqrt(x^2+1))#

I prefer to use the substitution

#x=sinhu#, #=>#, #dx=coshudu#

Therefore,

#int(x^3dx)/sqrt(1+x^2)=int(sinh^3ucoshudu)/(coshu)#

#=intsinh^3udu#

#=intsinh^2usinhudu#

#=int(cosh^2u-1)sinhudu#

Let #v=coshu#, #=>#, #dv=sinhudu#

Therefore,

#int(cosh^2u-1)sinhudu=int(v^2-1)dv#

#=1/3v^3-v#

#=1/3cosh^3u-coshu#

#=1/3cosh ^3(arcsinhx)-cosharcsinhx+C#

#cosh(arcsinh(x))=1/2(e^(arcsinhx)+e^(-arcsinhx))#

#=1/2(e^(ln(x+sqrt(x^2+1)))+e^(-ln(x+sqrt(x^2+1))))#

#=1/2((x+sqrt(x^2+1))+1/(x+sqrt(x^2+1)))#

#=1/2((x+sqrt(x^2+1))+(x-sqrt(x^2+1))/(x+sqrt(x^2+1)(x-sqrt(x^2+1))))#

#=1/2((x+sqrt(x^2+1))+(x-sqrt(x^2-1)/(x^2-x^2-1)))#

#=sqrt(x^2+1)#

And

#cosh^3(arcsinhx)=(sqrt(x^2+1))^3#

#=(x^2+1)sqrt(x^2+1)#

Finally,

#int(x^3dx)/sqrt(1+x^2)=1/3(x^2+1)sqrt(x^2+1)-sqrt(x^2+1)+C#

#=1/3(x^2-2)sqrt(x^2+1)+C#