Question

Pleas, how solve: Integrade 1/[x^2 Root(9+4x^2)] dx?

Answer 1

Use the substitution `2x=3tantheta`.

Explanation:

Let

`I=int1/(x^2sqrt(9+4x^2))dx`

Apply the substitution `2x=3tantheta`:

`I=int1/((9/4tan^2theta)(3sectheta))(3/2sec^2thetad theta)`

Simplify:

`I=2/9intcscthetacotthetad theta`

Integrate directly:

`I=-2/9csctheta+C`

Rewrite in terms of `3tantheta` and `3sectheta`:

`I=-2/9 (3sectheta)/(3tantheta)+C`

Reverse the substitution:

`I=-2/9sqrt(9+4x^2)/(2x)+C`

Simplify:

`I=-sqrt(9+4x^2)/(9x)+C`

Answer 2

`int1/(x^2sqrt(9+4x^2))dx=-sqrt(9+4x^2)/(9x)+C`

Explanation:

.

`int1/(x^2sqrt(9+4x^2))dx`

We need to use trigonometric substitution to solve this integral. Let's first set up our right triangle and write the three basic trigonometric functions for our angle `alpha`:

Assigning measurements of `3 and 2x` to the sides and using the Pythagorean formula gives us the measure of the hypotenuse:

`("Hypotenuse")^2=3^2+(2x)^2=9+4x^2`

`"Hypotenuse"=sqrt(9+4x^2`

`sinalpha=(2x)/sqrt(9+4x^2)`

`cosalpha=3/sqrt(9+4x^2)`

`tanalpha=(2x)/3`

`int1/(x^2sqrt(9+4x^2))dx=int(1/x^2)(1/sqrt(9+4x^2))(dx)`

We need to find each of the three pieces in the integral argument in terms of trigonometric functions in order to substitute them in.

From the tangent equation:

`3tanalpha=2x, :. x=(3tanalpha)/2, :. x^2=(9tan^2alpha)/4`

`1/x^2=4/(9tan^2alpha)`

From the cosine function:

`1/sqrt(9+4x^2)=cosalpha/3`

Taking the derivative of the tangent function, we get:

`3tanalpha=2x`

`3sec^2alphadalpha=2dx`

`dx=(3sec^2alphadalpha)/2`

Now, let's substitute:

`int1/(x^2sqrt(9+4x^2))dx=int(1/x^2)(1/sqrt(9+4x^2))(dx)=int(4/(9tan^2alpha))(cosalpha/3)((3sec^2alpha)/2)dalpha=int(2(2))/((9sin^2alpha)/cos^2alpha)(cosalpha/3)(3/(2cosalphacosalpha))dalpha=int(2cancel(color(red)((2))))/((9sin^2alpha)/cos^2alpha)(cancelcolor(green)(cosalpha)/cancelcolor(red)(3))(cancelcolor(red)(3)/(cancelcolor(red)(2)cancelcolor(green)(cosalpha)cosalpha))dalpha=`

`int(2cosalphacosalpha)/(9sin^2alpha)*1/cosalphadalpha=`

`int(2cancelcolor(red)(cosalpha)cosalpha)/(9sin^2alpha)*1/cancelcolor(red)(cosalpha)dalpha=int(2cosalpha)/(9sin^2alpha)dalpha=`

`2/9intcosalpha/sinalpha(1/sinalpha)dalpha=`

`2/9intcotalphacscalphadalpha=-2/9cscalpha+C`

Now, we can substitute back for `cscalpha`:

From the sine function above:

`cscalpha=1/sinalpha=1/((2x)/sqrt(9+4x^2))=sqrt(9+4x^2)/(2x)`

`int1/(x^2sqrt(9+4x^2))dx=-2/9*sqrt(9+4x^2)/(2x)+C=`

`-cancelcolor(red)(2)/9*sqrt(9+4x^2)/(cancelcolor(red)(2)x)+C=-sqrt(9+4x^2)/(9x)+C`