Integrate #intx^3/sqrt(x^2+4)# using trig substitution?

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Answer 1 · 2017-10-11T06:43:38

See the explanatiom below

You have to change as follows

#I=8(1/3u^3-u)#

#I=8/3(sec^3theta-3sectheta)#

#=8/3(((x^2+1)/2)^(3/2)-3sec(arctan(x/2))+C#

It's easier without trigonometric substitution

Let #u=x^2+4#, #=>#, #du=2xdx#

#I=1/2int((u-4)du)/sqrtu#

#=1/2intsqrtudu-int4/sqrtudu#

#=(u^(3/2)/3-4sqrtu)#

#=1/3(x^2+4)^(3/2)-4sqrt(x^2+4)#

#=((x^2-8))/3sqrt(x^2+4)+C#