# Another trig. sub

1. Feb 10, 2005

i dont know why i'm having so much problem finding a appropriate trigonometric sub for x. I can do the rest fine as long as i can find the right sub. please help me with the right trig. sub for this problem:

$$\int x*sqrt(6x^2-36x+49)dx$$

here's what i done:

$$\int 6x*sqrt(x^2-6x+\frac{49}{6})$$

then i did complete the square...

$$\int 6x* sqrt((x-3)^2 -5/6)$$

so i came up with x = sqrt(5/6)*sec(t) as my trig. sub, but it's incorrect. can someone lend me a hand?

2. Feb 10, 2005

### Kelvin

$$$\begin{gathered} \int_{}^{} {x\sqrt {6x^2 - 36x + 49} dx} \hfill \\ = \int_{}^{} {x\sqrt 6 \sqrt {x^2 - 6x + \frac{{49}} {6}} } dx \hfill \\ = \sqrt 6 \int_{}^{} {x\sqrt {\left( {x - 3} \right)^2 - \left( {\sqrt {\frac{5} {6}} } \right)^2 } } dx \hfill \\ \hbox{Put } x - 3 = \sqrt {\frac{5} {6}} \sec \theta \hfill \\ dx = \sqrt {\frac{5} {6}} \sec \theta \tan \theta d\theta \hfill \\ \sqrt 6 \int_{}^{} {\left( {\sqrt {\frac{5} {6}} \sec \theta + 3} \right)} \sqrt {\frac{5} {6}} \tan \theta \sqrt {\frac{5} {6}} \sec \theta \tan \theta d\theta \hfill \\ \hbox{I think you can do the rest (though this is not the fastest method)} \end{gathered}$$$