How do you integrate #int (x^2 -x) / sqrt(9 - x^2) dx# using trigonometric substitution?

1 Answer
maganbhai P.
May 13, 2018

#I=9/2sin^-1(x/3)-x/2sqrt(9-x^2)+sqrt(9-x^2)+c#

Explanation:

We know that,

#color(red)((1)sin^2x=(1-cos2x)/2)#

#color(blue)((2)intcosxdx=sinx+c)#

#color(violet)((3)intsinxdx=-cosx+c)#

Here,

#I=int(x^2-x)/sqrt(9-x^2)dx#

Subst. #x=3sinu=>dx=3cosudu#

#=>x^2=9sin^2u and sinu=x/3=>u=sin^-1(x/3)#

So,

#I=int(9sin^2u-3sinu)/sqrt(9-9sin^2u)xx3cosudu#

#=int(9sin^2u-3sinu)/(cancel(3cosu))xxcancel(3cosu)du#

#=9intsin^2udu-3intsinudu...tocolor(red)(Apply(1)#

#=9int(1-cos(2u))/2du-3intsinudu...toApplycolor(blue)((2)) andcolor(violet)((3))#

#=9/2[u-sin(2u)/2]-3(-cosu)+c#

#=9/4[2u-sin2u]+3cosu+c#

Subst. back , #sinu=x/3 and u=sin^-1(x/3)#

#I=9/4[2sin^-1(x/3)-2sinucosu]+3sqrt(1-sin^2u)+c#

#=9/4xx2[sin^-1 (x/3)-sinusqrt(1-sin^2u)]+3sqrt(1-x^2/9)+c#

#=9/2[sin^-1(x/3)-(x/3)sqrt(1-x^2/9)]+3sqrt(9-x^2)/3+c#

#=9/2sin^-1(x/3)-9/2xxx/3sqrt(9-x^2)/3+sqrt(9-x^2)+c#

#I=9/2sin^-1(x/3)-x/2sqrt(9-x^2)+sqrt(9-x^2)+c#

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