Question

How do you evaluate the integral `int sqrt(x^2-1)dx`?

Answer 1

`intsqrt(x^2-1)color(white).dx=1/2xsqrt(x^2-1)-1/2lnabs(x+sqrt(x^2-1))+C`

Explanation:

`I=intsqrt(x^2-1)color(white).dx`

Use the trigonometric substitution `x=sectheta`. Differentiating this shows that `dx=secthetatanthetacolor(white).d theta`. Substituting both of these gives:

`I=intsqrt(sec^2theta-1)(secthetatanthetacolor(white).d theta)`

Note that from the identity `tan^2theta+1=sec^2theta` we see that `tantheta=sqrt(sec^2theta-1)`:

`I=inttan(secthetatantheta)d theta`

`I=intsecthetatan^2thetacolor(white).d theta`

Let `tan^2theta=sec^2theta-1`:

`I=intsectheta(sec^2theta-1)color(white).d theta`

`I=intsec^3thetacolor(white).d theta-intsecthetacolor(white).d theta`

The integral of `sectheta` is common:

`I=intsec^3thetacolor(white).d theta-lnabs(sectheta+tantheta)`

Let `J=intsec^3thetacolor(white).d theta`.

`J=intsectheta(sec^2theta)d theta`

This can be tackled using integration by parts. Let:

`{(u=sectheta" "=>" "du=secthetatanthetacolor(white).d theta),(dv=sec^2thetacolor(white).d theta" "=>" "v=tantheta):}`

Then:

`J=secthetatantheta-intsecthetatan^2thetacolor(white).d theta`

Again let `tan^2theta=sec^2theta-1`:

`J=secthetatantheta-intsectheta(sec^2theta-1)d theta`

`J=secthetatantheta-intsec^3thetacolor(white).d theta+intsecthetacolor(white).d theta`

Note that the original integral `J` is included again, and we know the integral of secant:

`J=secthetatantheta-J+lnabs(sectheta+tantheta)`

`2J=secthetatantheta+lnabs(sectheta+tantheta)`

`J=1/2secthetatantheta+1/2lnabs(sectheta+tantheta)`

Returning to the original integral:

`I=J-lnabs(sectheta+tantheta)`

`I=(1/2secthetatantheta+1/2lnabs(sectheta+tantheta))-lnabs(sectheta+tantheta)`

`I=1/2secthetatantheta-1/2lnabs(sectheta+tantheta)`

Recall our original substitution `x=sectheta`. This implies that `tantheta=sqrt(sec^2theta-1)=sqrt(x^2-1)`.

`I=1/2xsqrt(x^2-1)-1/2lnabs(x+sqrt(x^2-1))+C`

Answer 2

`x/2sqrt(x^2-1)-1/2ln|x+sqrt(x^2-1)|+C`.

Explanation:

Let `I=intsqrt(x^2-1)dx=int(sqrt(x^2-1))(1)dx`

We will use the following Rule of Integration by Parts (IBP) :

`IBP : intuvdx=uintvdx-int{(du)/dxintvdx)}dx`

Taking `u=sqrt(x^2-1) rArr (du)/dx=1/(2sqrt(x^2-1))d/dx(x^2-1), i.e.,`

`(du)/dx=x/sqrt(x^2-1)`

`v=1 rArr intvdx=x` Therefore,

`I=xsqrt(x^2-1)-int{(x/sqrt(x^2-1))(x)}dx`

`=xsqrt(x^2-1)-intx^2/sqrt(x^2-1)dx`

`=xsqrt(x^2-1)-int{(x^2-1)+1}/sqrt(x^2-1)dx`

`=xsqrt(x^2-1)-int{(x^2-1)/sqrt(x^2-1)+1/sqrt(x^2-1)}dx`

`=xsqrt(x^2-1)-intsqrt(x^2-1)dx-int1/sqrt(x^2-1)dx, i.e.,`

`I=xsqrt(x^2-1)-I-ln|x+sqrt(x^2-1)|`

`rArr I+I=2I=xsqrt(x^2-1)-ln|x+sqrt(x^2-1)|," and, therefore,"`

`I=x/2sqrt(x^2-1)-1/2ln|x+sqrt(x^2-1)|+C`.

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