Question

How do you integrate `int 1/(x^2+4)^(3/2)` by trigonometric substitution?

Answer 1

`intdx/(x^2+4)^(3/2)=1/4x/sqrt(x^2+4)+C`

Explanation:

Rewrite with the rational exponent converted to a root:

`intdx/(sqrt((x^2+4)^3)`

Let `x=2tantheta`
`dx=2sec^2thetad theta`

Rewrite:

`int(2sec^2theta)/(sqrt((4tan^2theta+4)^3))d theta`

`int(2sec^2theta)/(sqrt((4(tan^2theta+1))^3))d theta`

Recall the identity

`tan^2theta+1=sec^2theta`, and apply it:

`int(2sec^2theta)/(sqrt(64(sec^2theta)^3))d theta`

`1/4intsec^2theta/sqrt(sec^6theta)d theta`

`1/4intsec^2theta/sec^3thetad theta`

`1/4intcosthetad theta=1/4sintheta+C`

We want to rewrite in terms of `x.` Recalling that `x=2tantheta, tantheta=x/2.` If `tantheta=x/2, sintheta=x/sqrt(x^2+4)`.

This could be deduced by drawing a right triangle with the sides opposite and adjacent to angle `theta` being labeled `x, 2` (respectively), making the hypotenuse `sqrt(x^2+4)`, and `sintheta=(opposite)/(hypoten use)=x/sqrt(x^2+4)`

So,

`intdx/(x^2+4)^(3/2)=1/4x/sqrt(x^2+4)+C`

Answer 2

`I=1/4*x/sqrt(x^2+4)+c`

Explanation:

Here,

`I=int1/(x^2+4)^(3/2)dx`

Let, `x=2tanu=>dx=2sec^2udu`

`andx^2+4=4tan^2u+4=4(tan^2u+1)=4sec^2u`

So,

`I=int(2sec^2u)/((4sec^2u)^(3/2))du=int(2sec^2u)/(2secu)^3du`

`=>I=int(2sec^2u)/(8sec^3u)du`

`=>I=int1/(4secu)du`

`=>I=1/4intcosdu`

`=>I=1/4sinu+c`

`=>I=1/2(sinu/cosu)cosu+c`

`=1/4tanu/secu+c`

`=1/4tanu/sqrt(tan^2u+1)+c`

`=1/4(x/2)/sqrt((x^2/4)+1)+c`

`=1/4*x/sqrt(x^2+4)+c`