I need some help with trig substituion

[ijd5000]

Hi,

I'm doing something wrong here, any suggestions?

i'm a bit rusty with completing the square, so the error is most likely there

 

[Deleted member 4993]

Hi,

I'm doing something wrong here, any suggestions?

i'm a bit rusty with completing the square, so the error is most likely there

15 + 6x - 9x² = 16 - (3x -1)² = 4² - (3x -1)²

Let

3x - 1 = 4 * sin(Θ)

dx = 4/3 * cos (Θ) dΘ

and continue.....

 

[ijd5000]

did it again and got

---

1/48[ (27x+23/(15+6x-9x²)^1/2 - 8sin^-1((3x-1)/4) ]+C

still wrong

 

[Deleted member 4993]

did it again and got

---

1/48[ (27x+23/(15+6x-9x²)^1/2 - 8sin^-1((3x-1)/4) ]+C

still wrong

I get:

I(Θ) = 1/(27*16) * [ 17*tan(Θ) - 8*Θ + 8*sec(Θ) ] + C ...................... edited

 

[ijd5000]

I get:

I(Θ) = 1/(27*16) * [ 17*tan(Θ) - 8*Θ + 8*sec(Θ) ] + C ...................... edited

did it again i get 1/27[17 tan(theta) +8 sec(theta)-16(theta)]+C

where is that factor of 16 coming from? Also the coefficient in front of theta?