Question

Question #34e5a

Answer 1

`lnabs(x+sqrt(x^2-9))-x/sqrt(x^2-9)+C`

Explanation:

`I=intx^2/(x^2-9)^(3/2)dx`

Use the substitution `x=3sectheta`. This implies that `dx=3secthetatanthetad theta`. Importantly, `x^2-9=9sec^2theta-9=9(sec^2theta-1)=9tan^2theta`.

`I=int(9sec^2theta)/(9tan^2theta)^(3/2)(3secthetatanthetad theta)`

`I=int(27sec^3thetatantheta)/(27tan^3theta)d theta`

`I=intsec^3theta/tan^2thetad theta`

`I=int1/cos^3thetacos^2theta/sin^2theta`

`I=intcsc^2thetasecthetad theta`

Use `csc^2theta=cot^2theta+1`:

`I=int(cot^2theta+1)secthetad theta`

`I=int(cos^2theta/sin^2theta 1/costheta+sectheta)d theta`

`I=int(cotthetacsctheta+sectheta)d theta`

These have common integrals:

`I=-csctheta+lnabs(sectheta+tantheta)+C`

Our original substitution was `sectheta=x/3`. This means that `costheta=3/x`, so we have a right triangle where the side adjacent to `theta` is `3` and the hypotenuse is `x`. Through the Pythagorean Theorem, the side opposite `theta` is `sqrt(x^2-9)`. Then:

• `tantheta="opp"/"adj"=sqrt(x^2-9)/3`
• `csctheta=1/sintheta="hyp"/"opp"=x/sqrt(x^2-9)`

Hence:

`I=lnabs(x/3+sqrt(x^2+9)/3)-x/sqrt(x^2-9)+C`

Factor `1/3` from the natural logarithm. Using `log(AB)=log(A)+log(B)`, a `ln(1/3)` will be spit out of the logarithm but it can be neglected because as a constant it will be absorbed into `C`, the constant of integration.

`I=lnabs(x+sqrt(x^2-9))-x/sqrt(x^2-9)+C`