# trignometric substitution

• Feb 1st 2009, 05:29 PM
twilightstr
trignometric substitution
integrate integral of 1/(square root of 9x^2+6x-8)dx.
my work: integral dx/(suare root 9(x+ (1/3))^2 -7
u= x+(1/3) du=dx
integral of du/(9u^2-7) u=square root 7/3 sec theta squareroot (7sectheta^2 - 7) =square root 7theta^2.
need help? i dont think i dont thing i did any of this right.
• Feb 1st 2009, 05:43 PM
Il Conte di Libenskof
Well, I would suggest first to complete the square. You should have 9*x^2 + 6*x - 8 = 9*x^2 + 6*x + 1 - 9 = (3*x + 1)^2 - 9. Then I suggest to make a little triangle - that's how I usually did it - or just implement those little formulas you're given in your textbook. But if I make a triangle, I denote the hypotenuse as (3*x + 1), the angle U, the adjacent side = 3. so, sec(u) = (3*x + 1)/3, so (3*x+1) = 3*sec(u). You substitute your (3*x+1) back, and do not forget, that x = sec(u) - 1/3. Then dx = sec(u)*tan(ua)du. So, your new integral is (1/sqrt(9*(sec(u))^2-9))*sec(u)*tan(u)du. Try to do this!