Question

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

Answer 1

`-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`.

Answer 2

`-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!