How do you integrate #int -x^3/sqrt(144+x^2)dx# using trigonometric substitution?

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Answer 1 · 2018-07-06T12:28:33

The answer is #=-1/3(144+x^2)^(3/2)+144(144+x^2)^(1/2)+C#

There is no need for trigonometric substitution

The integral is

#I=int(-x^3dx)/sqrt(144+x^2)=-int(x^3dx)/sqrt(144+x^2)#

Let #u=144+x^2#, #=>#, #du=2xdx#

Therefore,

#I=-1/2int((u-144)du)/sqrtu#

#=-1/2int(sqrtu-144/sqrtu)du#

#=-1/2(u^(3/2)/(3/2)-(144u^(1/2))/(1/2))#

#=-1/3u^(3/2)+144u^(1/2)#

#=-1/3(144+x^2)^(3/2)+144(144+x^2)^(1/2)+C#