Question

How do you integrate `int x/sqrt(3x^2-6x-8) dx` using trigonometric substitution?

Answer 1

`intx/sqrt(3x²-6x-8)dx=sqrt11/3cot(csc^(-1)(sqrt(3/11)(x-1)))+1/sqrt3ln(|cot(csc^(-1)(sqrt(3/11)(x-1)))+sqrt(3/11)(x-1)|)+C`, `C in RR`

Explanation:

`intx/sqrt(3x²-6x-8)dx`

`=1/sqrt(3)intx/sqrt(x²-2x-8/3)dx`

`=1/sqrt(3)intx/sqrt((x-1)²-11/3)`

Let `x-1=sqrt(11/3)csc(theta)`

`dx=-sqrt(11/3)cot(theta)csc(theta)d theta`

`1/sqrt(3)intx/sqrt((x-1)²-11/3)=-1/sqrt(3)int((sqrt(11/3)csc(theta)+1)cot(theta)csc(theta))/sqrt(csc²(theta)-1)d theta`
Because `csc²(x)-1=cot²(x)`,
`=-1/sqrt(3)(intsqrt(11/3)csc²(theta)d theta + intcsc(theta)d theta)`
`=sqrt11/3cot(theta)+1/sqrt3ln(|cot(theta)+csc(theta)|)`
Finally, `theta=csc^(-1)(sqrt(3/11)(x-1))`
So:
`intx/sqrt(3x²-6x-8)dx=sqrt11/3cot(csc^(-1)(sqrt(3/11)(x-1)))+1/sqrt3ln(|cot(csc^(-1)(sqrt(3/11)(x-1)))+sqrt(3/11)(x-1)|)+C`, `C in RR`
\0/ here's our answer !