How do you integrate #int x^2sqrt(3-x)# using substitution?

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Answer 1 · 2016-08-14T07:13:09

#-2/35(3-x)^(3/2)(24+12x+5x^2)+C#.

To get rid of the Radical Sign, i.e., the Square Root, let us subst.

#3-x=t^2, or, x=3-t^2, so, dx=-2tdt#. Hence,

#I=intx^2sqrt(3-x)dx=int(3-t^2)^2*t*(-2t)dt#

#=-2int(9-6t^2+t^4)t^2dt#

#=-2int(9t^2-6t^4+t^6)dt#

#=-18intt^2dt+12intt^4dt-2intt^6dt#

#=-18*t^3/3+12*t^5/5-2*t^7/7#

#=-6t^3+12/5t^5-2/7t^7#

#=-2/35*t^3(105-42t^2+5t^4)#

#=-2/35*(t^2)^(3/2)(105-42t^2+5t^4)#

#=-2/35(3-x)^(3/2){105-42(3-x)+5(3-x)^2}#

#=-2/35(3-x)^(3/2)(24+12x+5x^2)+C#.

Enjoy Maths.!