How do you integrate $\int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}}$ $int 1/sqrt(9x^2+6x-8)$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}}$ $int 1/sqrt(9x^2+6x-8)$ by trigonometric substitution?

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Answer 1 · 2017-02-15T19:41:39

The answer is $\frac{1}{3} ln ∣ ∣ ∣ ∣ x + \frac{1}{3} + \sqrt{{(x + \frac{1}{3})}^{2} - 1} ∣ ∣ ∣ ∣ + C$ $1/3ln|x +1/3 + sqrt((x + 1/3)^2 - 1)| + C$ . See below for details.

Explanation:

Complete the square under the square root.

$\int \frac{1}{\sqrt{9 (x^{2} + \frac{2}{3} x + \frac{1}{9} - \frac{1}{9}) - 8}} d x$ $int 1/sqrt(9(x^2 + 2/3x + 1/9 - 1/9)- 8)dx$

$\int \frac{1}{\sqrt{9 {(x + \frac{1}{3})}^{2} - 1 - 8}} d x$ $int 1/sqrt(9(x + 1/3)^2 - 1 - 8)dx$

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

Now let $u = x + \frac{1}{3}$ $u = x + 1/3$ . Then $d u = d x$ $du = dx$ .

$\int \frac{1}{\sqrt{9 u^{2} - 9}} d u$ $int 1/sqrt(9u^2 - 9)du$

$\int \frac{1}{\sqrt{9 (u^{2} - 1)}} d u$ $int 1/sqrt(9(u^2 - 1))du$

We are of the form $\sqrt{x^{2} - a^{2}}$ $sqrt(x^2- a^2)$ . This means using the substitution $x = a sec θ$ $x = asectheta$ . Now, let $u = sec θ$ $u = sectheta$ . Then $d u = sec θ tan θ d θ$ $du = secthetatantheta d theta$ .

$\int \frac{1}{\sqrt{9 ({sec}^{2} θ - 1)}} \cdot sec θ tan θ d θ$ $int 1/sqrt(9(sec^2theta - 1)) * secthetatantheta d theta$

$\int \frac{1}{\sqrt{9 {tan}^{2} θ}} \cdot sec θ tan θ d θ$ $int 1/sqrt(9tan^2theta) * sec thetatantheta d theta$

$\int \frac{1}{3 tan θ} \cdot sec θ tan θ d θ$ $int 1/(3tantheta) * secthetatantheta d theta$

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

This is a relatively well known integral which can be derived here

$\frac{1}{3} ln | sec θ + tan θ | + C$ $1/3ln|sectheta + tantheta| + C$

Recall the original trig substitution was $\frac{u}{1} = sec θ$ $u/1 = sectheta$ . This means that the side opposite $θ$ $theta$ measures $\sqrt{u^{2} - 1}$ $sqrt(u^2 - 1)$ . Therefore, $tan θ = \frac{\sqrt{u^{2} - 1}}{1} = \sqrt{u^{2} - 1}$ $tantheta = sqrt(u^2 - 1)/1 = sqrt(u^2 - 1)$ .

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

Now reverse the other substitution.

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

Hopefully this helps!

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How do you integrate $\int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}}$ $int 1/sqrt(9x^2+6x-8)$ by trigonometric substitution?

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How do you integrate ∫1√9x2+6x−8int 1/sqrt(9x^2+6x-8) by trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}}$ $int 1/sqrt(9x^2+6x-8)$ by trigonometric substitution?