# How do you integrate int 1/sqrt(3x-12sqrtx+29)  using trigonometric substitution?

Jul 16, 2018

$= \frac{2}{3} \left(\sqrt{3 x - 12 \sqrt{x} + 29} + 2 \sqrt{3} {\sinh}^{-} 1 \left(\sqrt{\frac{3}{17}} \left(\sqrt{x} - 2\right)\right)\right) + C$

#### Explanation:

For the integrand $\frac{1}{\sqrt{3 x - 12 \sqrt{x} + 29}}$, substitute

$u = \sqrt{x}$ and $\mathrm{du} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$

$= 2 \int \left(\frac{u}{\sqrt{3 {u}^{2} - 12 u + 29}}\right) \mathrm{du}$

Completing the square, we get

$= 2 \int \left(\frac{u}{\sqrt{{\left(\sqrt{3} u - 2 \sqrt{3}\right)}^{2} + 17}}\right) \mathrm{du}$

Let $s = \sqrt{3} u - 2 \sqrt{3}$ and $\mathrm{ds} = \sqrt{3} \mathrm{du}$

$= \frac{2}{\sqrt{3}} \int \left(\frac{s + 2 \sqrt{3}}{\sqrt{3} \sqrt{{s}^{2} + 17}} \mathrm{ds}\right)$

Factor out the constant $\frac{1}{\sqrt{3}}$

$= \frac{2}{3} \int \frac{s + 2 \sqrt{3}}{\sqrt{{s}^{2} + 17}} \mathrm{ds}$

Next, substitute $s = \sqrt{17} \tan \left(p\right)$ and $\mathrm{ds} = \sqrt{17} {\sec}^{2} \left(p\right) \mathrm{dp}$

Then, $\sqrt{{s}^{2} + 17} = \sqrt{17 {\tan}^{2} \left(p\right) + 17} = \sqrt{17} \sec \left(p\right)$ and $p = {\tan}^{-} 1 \left(\frac{s}{\sqrt{17}}\right)$. This gives us

$= \frac{2 \sqrt{17}}{3} \int \frac{\sqrt{17} \tan \left(p\right) + 2 \sqrt{3} \sec \left(p\right)}{\sqrt{17}} \mathrm{dp}$

Factoring out $\frac{1}{\sqrt{17}}$ gives

$= \frac{2}{3} \int \left(\sqrt{17} \tan \left(p\right) + 2 \sqrt{3}\right) \sec \left(p\right) \mathrm{dp}$

$= \frac{2}{3} \int \left(\sqrt{17} \tan \left(p\right) \sec \left(p\right) + 2 \sqrt{3} \sec \left(p\right)\right) \mathrm{dp}$

=(2sqrt(17))/3 int (sqrt(17)tan(p)sec(p))dp + 4/sqrt(3) int sec(p)) dp

Substitute $w = \sec \left(p\right)$ and $\mathrm{dw} = \tan \left(p\right) \sec \left(p\right) \mathrm{dp}$

$= \frac{2 \sqrt{17}}{3} \int 1 \mathrm{dw} + \frac{4}{\sqrt{3}} \int \sec \left(p\right) \mathrm{dp}$

$= \frac{2 \sqrt{17} w}{3} + \frac{4}{\sqrt{3}} \int \sec \left(p\right) \mathrm{dp}$

$= \frac{4 \ln \left(\tan \left(p\right) + \sec \left(p\right)\right)}{\sqrt{3}} + \frac{2 \sqrt{17} 3}{3} + C$

Backwards substituting all the variables eventually gives

$= \frac{2}{3} \left(\sqrt{3 x - 12 \sqrt{x} + 29} + 2 \sqrt{3} {\sinh}^{-} 1 \left(\sqrt{\frac{3}{17}} \left(\sqrt{x} - 2\right)\right)\right) + C$