Use an appropriate substitution and then a trigonometric substitution to evaluate the following integral? #int \ x^3/(sqrt(49+x^8)) \ dx#

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Answer 1 · 2018-05-23T06:33:47

The answer is #=1/4ln(sqrt(1+x^8/49)+x^4/7)+C#

Let #u=x^4/7#

#=>#, #du=4/7x^3dx#

The integral is

#I=int(x^3dx)/sqrt(49+x^8)=7/4int(du)/sqrt(49+49u^2)#

#=1/4int(du)/sqrt(u^2+1)#

Let #u=tanv#, #=>#, #du=sec^2vdv#

#1+tan^2v=sec^2v#

The integral becomes

#I=1/4int(sec^2vdv)/sqrt(1+tan^2v)#

#=1/4intsecvdv#

#=1/4int(secv(secv+tanv)dv)/(secv+tanv)#

#=1/4int((sec^2v+secvtanv)dv)/(secv+tanv)#

Let #w=secv+tanv#

#=>#, #dw=secvtanv+sec^2v#

The integral is

#I=1/4int(dw)/(w)#

#=1/4lnw#

#=1/4ln(secv+tanv)#

#=1/4ln(sqrt(1+u^2)+u)#

#=1/4ln(sqrt(1+x^8/49)+x^4/7)+C#