How do you integrate $\int \frac{1}{\sqrt{9 x^{2} - 6 x + 2}}$ $int 1/sqrt(9x^2-6x+2)$ using trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{9 x^{2} - 6 x + 2}}$ $int 1/sqrt(9x^2-6x+2)$ using trigonometric substitution?

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Answer 1 · 2018-03-13T08:55:37

The answer is $= \frac{1}{3} ln (∣ ∣ ∣ (3 x - 1) + \sqrt{{(3 x - 1)}^{2} + 1} ∣ ∣ ∣) + C$ $=1/3ln(|(3x-1)+sqrt((3x-1)^2+1)|)+C$

Explanation:

Let complete the square in the denominator

$9 x^{2} - 6 x + 2 = 9 x^{2} - 6 x + 1 + 2 - 1 = {(3 x - 1)}^{2} + 1$ $9x^2-6x+2=9x^2-6x+1+2-1=(3x-1)^2+1$

Perform the substitution

$u = 3 x - 1$ $u=3x-1$ , $\Rightarrow$ $=>$ , $d u = 3 d x$ $du=3dx$

Therefore,

$\int \frac{d x}{\sqrt{9 x^{2} - 6 x - 2}} = \frac{1}{3} \int \frac{d u}{\sqrt{u^{2} + 1}}$ $int(dx)/(sqrt(9x^2-6x-2))=1/3int(du)/sqrt(u^2+1)$

Let $u = tan θ$ $u=tantheta$ , $\Rightarrow$ $=>$ , $d u = {sec}^{2} θ d θ$ $du=sec^2thetad theta$

Therefore,

$\int \frac{d x}{\sqrt{9 x^{2} - 6 x - 2}} = \frac{1}{3} \int \frac{{sec}^{2} θ d θ}{sec θ}$ $int(dx)/(sqrt(9x^2-6x-2))=1/3int(sec^2thetad theta)/(sectheta)$

$= \frac{1}{3} \int sec θ d θ$ $=1/3intsecthetad theta$

$= \frac{1}{3} \int \frac{sec θ (sec θ + tan θ) d θ}{sec θ + tan θ}$ $=1/3int(sectheta(sectheta+tantheta)d theta)/(sectheta+tantheta)$

Let $v = sec θ + tan θ$ $v=sectheta+tantheta$

$d v = ({sec}^{2} θ + sec θ tan θ) d θ$ $dv=(sec^2theta +secthetatantheta)d theta$

So,

$\int \frac{d x}{\sqrt{9 x^{2} - 6 x - 2}} = \frac{1}{3} \int \frac{d v}{v}$ $int(dx)/(sqrt(9x^2-6x-2))=1/3int(dv)/v$

$= \frac{1}{3} ln v$ $=1/3lnv$

$= \frac{1}{3} ln (sec θ + tan θ)$ $=1/3ln(sectheta+tantheta)$

$= \frac{1}{3} ln (u + \sqrt{u^{2} + 1})$ $=1/3ln(u+sqrt(u^2+1))$

$= \frac{1}{3} ln (∣ ∣ ∣ (3 x - 1) + \sqrt{{(3 x - 1)}^{2} + 1} ∣ ∣ ∣) + C$ $=1/3ln(|(3x-1)+sqrt((3x-1)^2+1)|)+C$

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How do you integrate $\int \frac{1}{\sqrt{9 x^{2} - 6 x + 2}}$ $int 1/sqrt(9x^2-6x+2)$ using trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{9 x^{2} - 6 x + 2}}$ $int 1/sqrt(9x^2-6x+2)$ using trigonometric substitution?