How do you integrate #int x/sqrt(3x^2-6x+10) dx# using trigonometric substitution?

Question

William Abdalla · Accepted Answer

#int x/sqrt(3x^2-6x+10) dx=1/12sqrt(3x^2-6x+10)+6/sqrt7ln|(1/7(3x^2-6x+10))+sqrt(3/7)(x-1)|#

#int x/sqrt(3x^2-6x+10) dx#

= #1/6int(6x-6)/sqrt(3x^2-6x+10)dx+6/sqrt3int(dx)/sqrt(x^2-2x+10/3)#

For the first part let #u=3x^2-6x+10# ten #du=(6x-6)dx#

and integral becomes #1/6int(du)/(sqrtu)=1/6*1/2*u^(1/2)#

= #1/12sqrt(3x^2-6x+10)#

Second part can be written as

#2sqrt3int1/sqrt((x-1)^2+7/3)dx=#

Now let #sqrt(3/7)(x-1)=tant# then #sqrt(3/7)dx=sec^2tdt# and our integral becomes

#2sqrt3sqrt(3/7)intsec^2t/sectdt=6/sqrt7ln|sect+tant|#

= #6/sqrt7ln|sqrt(3/7(x-1)^2+1)+sqrt(3/7)(x-1)|#

= #6/sqrt7ln|(1/7(3x^2-6x+10))+sqrt(3/7)(x-1)|#

and hence #int x/sqrt(3x^2-6x+10) dx#

= #1/12sqrt(3x^2-6x+10)+6/sqrt7ln|(1/7(3x^2-6x+10))+sqrt(3/7)(x-1)|#

Isaiah Allen · Answer

undefined To integrate $\frac{x}{\sqrt{3x^2-6x+10}}$ using trigonometric substitution, follow these steps:

1. Complete the square under the radical to get $3x^2 - 6x + 10 = 3(x^2 - 2x + \frac{10}{3}) = 3((x - 1)^2 + \frac{1}{3})$.

2. Perform the substitution $x - 1 = \frac{\sqrt{3}}{3} \tan(\theta)$.

3. Calculate $dx$ using this substitution: $dx = \frac{\sqrt{3}}{3} \sec^2(\theta) d\theta$.

4. Substitute $x$ and $dx$ into the integral.

5. Simplify the integral in terms of $\theta$.

6. Integrate with respect to $\theta$.

7. Finally, substitute back $x$ in terms of $\theta$ and simplify.

The final integral should be in a form that can be easily solved using trigonometric techniques.