How do you integrate #int -x^3/sqrt(9+9x^2)dx# using trigonometric substitution?

2 Answers
May 4, 2018

# -1/9(x^2-2)sqrt(x^2+1)+C#.

Explanation:

We have, #int-x^3/sqrt(9+9x^2)dx=-1/3intx^3/sqrt(1+x^2)dx#.

Let, #I=intx^3/sqrt(1+x^2)dx#.

Subst. #x=tany," so that, "dx=sec^2ydy#.

#:. I=inttan^3y/sqrt(1+tan^2y)sec^2ydy#,

#=inttan^3y/secysec^2ydy#,

#=inttan^3ysecydy#,

#=inttan^2y*tanysecydy#,

#=int(sec^2y-1)tanysecydy#.

Next we subst. #secy=t. :. secytanydy=dt#.

#:. I=int(t^2-1)dt#,

#=t^3/3-t#,

#=sec^3y/3-secy............[because, t=secy]#,

#=1/3secy(sec^2y-3)#,

#=1/3sqrt(sec^2y){(1+tan^2y)-3}#.

#=1/3sqrt(1+tan^2y)(tan^2y-2)#.

#rArr I=1/3(x^2-2)sqrt(x^2+1)......[because, tany=x]#.

Hence, #int-x^3/sqrt(9+9x^2)dx=-1/9(x^2-2)sqrt(x^2+1)+C#.

May 4, 2018

# -1/9(x^2-2)sqrt(x^2+1)+C#.

Explanation:

I have solved the Problem using trigo. substn., but the same

can be done without it, as shown below :

#I=int-x^3/sqrt(9+9x^2)dx#,

#=-1/3int(x^2*x)/sqrt(x^2+1)dx#,

#=-1/3int[{(x^2+1)-1}x]/sqrt(x^2+1)dx#,

#=-1/3int{(x^2+1)/sqrt(x^2+1)-1/sqrt(x^2+1)}xdx#,

#=-1/3int{sqrt(x^2+1)-1/sqrt(x^2+1)}xdx#,

#=-1/3*1/2int{sqrt(x^2+1)-1/sqrt(x^2+1)}(2x)dx#,

#=-1/6int{(x^2+1)^(1/2)-(x^2+1)^(-1/2)}d(x^2+1)#,

#=-1/6{(x^2+1)^(1/2+1)/(1/2+1)-(x^2+1)^(-1/2+1)/(-1/2+1)}#,

#=-1/6{2/3(x^2+1)^(3/2)-2(x^2+1)^(1/2)}#,

#=-1/6*2/3(x^2+1)^(1/2){(x^2+1)^(3/2-1/2)-3(x^2+1)^(1/2-1/2)}#,

#=-1/9(x^2+1)^(1/2){(x^2+1)-3}#,

#=-1/9(x^2-2)sqrt(x^2+1)+C#, as before!