Trig substitution for ∫x√(6x²-36x+49)dx

[ProBasket]

i don't 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?

 

[Kelvin]

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

 

[wais]

It looks like you are on the right track with completing the square, but you may have made a small error in your substitution. Instead of using x = sqrt(5/6) * sec(t), try using x = sqrt(5/6) * tan(t). This should give you the correct trigonometric substitution. Remember to also adjust the limits of integration when using a trigonometric substitution. I hope this helps!