Question

How do you integrate `int x^2/sqrt(4x^2+25)dx` using trigonometric substitution?

Answer 1

`1/20(sqrt(4x+25))x - 5/8ln(sqrt(4x+25)/5+2/5x)`

Explanation:

`intx^2/sqrt(4x+25)dx`

`Let x = 5/2 tan A, dx = (5/2 sec^2a)dA`

`=intx^2/sqrt(4x+25)dx = int (5/2)^2(tan^2A sec^2A) /(sqrt(4(5/2)^2tan^2A+25)` `dA`

`=intx^2/sqrt(4x+25)dx = int (5/2)^2(tan^2A sec^2A) /(sqrt(25tan^2A+25)` `dA`

`=intx^2/sqrt(4x+25)dx = int (5/2)^2(tan^2A sec^2A) /5(sqrt(tan^2A+1))` `dA`

`sec^2A-1= tan^2A`

`= int (5/4)(tan^2A sec^2A) /(sqrt(sec^2A))` `dA`

`= int (5/4)((sec^2A-1) sec^2A) /(sqrt(sec^2A))` `dA`
`= int (5/4)((sec^3A-secA))` `dA`
`= (5/4)(int(sec^3A)-int(secA))` `dA` -------- equation (1)

We need to find

`int(secA)` `dA` = `int sec A(secA + tanA)/(secA+tanA)` `dA`

=`int sec A(sec^2A + tanAsecA)/(secA+tanA)` `dA`

Let `B= secA+tanA`
`dB` `= (secAtanA+sec^2A)` `dA`
`=int1/B` `dB`
`=ln |B|+c`
`=ln |secA+tanA|+c` -----equation 2

We now need to find

`int(sec^3A)` `dA`

Integrate by parts

`int sec^2AsecA` `dA`
Let `C= secA`
`dC` `= secAtanA` `dA`

`dD` `= sec^2A` `dA`
`D= int sec^2A` `dA` `=tanA + const`

Using by parts gives
`=int sec^2AsecA` `dA` `= CD - int D` `dC`/`dA`
`=secAtanA - int secAtanAtanA` `dA`
`=secAtanA - int (sec^2A-1)secA` `dA`
`=secAtanA - int (sec^3A-secA)` `dA`
`int sec^3AsecA` `dA` `=secAtanA - (int sec^3A -int secA)` `dA`

substituting using equation 2
`2int sec^3A` `dA` = `secAtanA+ ln|secA+tanA)`
`1/2(secAtanA+ ln|secA+tanA))` ----equation 3

From equation 1 we had

`intx^2/sqrt(4x+25)dx` `= (5/4)(int(sec^3A)-int(secA))` `dA`

using the results from equations 2 and 3

`=5/4((1/2(secAtanA + ln|(SecA+tanA)|))-ln|(SecA+tanA)|)`
`=5/8(secAtanA + -ln|(SecA+tanA)|))`

Now convert everything back to x
`x=5/2 tan A`, `tan A=2/5x`, `tan^2A=(2/5)^2x^2`
`sec^2A = tan^2A +1 =(4x^2+25)/25`
`secA = =sqrt((4x^2+25)/(25))= sqrt(4x+25)/5`

`1/20(sqrt(4x+25))x - 5/8ln(sqrt((4x+25))/(5)+2/5x)`