How do you integrate $\int \frac{1}{\sqrt{4 x + 8 \sqrt{x} + 12}}$ $int 1/sqrt(4x+8sqrtx+12)$ using trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{4 x + 8 \sqrt{x} + 12}}$ $int 1/sqrt(4x+8sqrtx+12)$ using trigonometric substitution?

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Answer 1 · 2016-09-07T23:50:26

$\sqrt{x + 2 \sqrt{x} + 3} - ln (∣ ∣ \sqrt{x} + \sqrt{x + 2 \sqrt{x} + 3} + 1 ∣ ∣) + C$ $sqrt(x+2sqrtx+3)-ln(abs(sqrtx+sqrt(x+2sqrtx+3)+1))+C$

Explanation:

We have:

$\int \frac{d x}{\sqrt{4 x + 8 \sqrt{x} + 12}}$ $intdx/sqrt(4x+8sqrtx+12)$

Factor $\sqrt{4} = 2$ $sqrt4=2$ from the denominator:

$= \frac{1}{2} \int \frac{d x}{\sqrt{x + 2 \sqrt{x} + 3}}$ $=1/2intdx/sqrt(x+2sqrtx+3)$

Complete the square in the denominator. Think about it in terms of $x^{2} + 2 x + 3$ $x^2+2x+3$ , and then switch $x$ $x$ terms to $\sqrt{x}$ $sqrtx$ and $x^{2}$ $x^2$ to $x$ $x$ :

$= \frac{1}{2} \int \frac{d x}{\sqrt{{(\sqrt{x} + 1)}^{2} + 2}}$ $=1/2intdx/sqrt((sqrtx+1)^2+2)$

Now, let $\sqrt{x} + 1 = \sqrt{2} tan θ$ $sqrtx+1=sqrt2tantheta$ . Note that this implies that $\frac{d x}{2 \sqrt{x}} = \sqrt{2} {sec}^{2} θ d θ$ $dx/(2sqrtx)=sqrt2sec^2thetad theta$ .

Rearranging some:

$= \int \frac{\sqrt{x} d x}{2 \sqrt{x} \sqrt{{(\sqrt{x} + 1)}^{2} + 2}}$ $=int(sqrtxdx)/(2sqrtxsqrt((sqrtx+1)^2+2))$

Now, performing the substitutions. We have $\frac{d x}{2 \sqrt{x}} = \sqrt{2} {sec}^{2} θ d θ$ $dx/(2sqrtx)=sqrt2sec^2thetad theta$ present, as well as $\sqrt{x} = \sqrt{2} tan θ - 1$ $sqrtx=sqrt2tantheta-1$ in the numerator. Don't forget the switch in the radical in the denominator either:

$= \int \frac{(\sqrt{2} tan θ - 1) \sqrt{2} {sec}^{2} θ d θ}{\sqrt{2 {tan}^{2} θ + 2}}$ $=int((sqrt2tantheta-1)sqrt2sec^2thetad theta)/sqrt(2tan^2theta+2)$

Factoring $\sqrt{2}$ $sqrt2$ from the denominator and using $\sqrt{{tan}^{2} θ + 1} = sec θ$ $sqrt(tan^2theta+1)=sectheta$ :

$= \int \frac{(\sqrt{2} tan θ - 1) \sqrt{2} {sec}^{2} θ}{\sqrt{2} sec θ} d θ$ $=int((sqrt2tantheta-1)sqrt2sec^2theta)/(sqrt2sectheta)d theta$

$= \int (\sqrt{2} tan θ - 1) sec θ d θ$ $=int(sqrt2tantheta-1)secthetad theta$

$= \sqrt{2} \int tan θ sec θ d θ - \int sec θ d θ$ $=sqrt2inttanthetasecthetad theta-intsecthetad theta$

Both of which are common integrals:

$= \sqrt{2} sec θ - ln (| sec θ + tan θ |)$ $=sqrt2sectheta-ln(abs(sectheta+tantheta))$

Write $sec θ$ $sectheta$ in terms of $tan θ$ $tantheta$ :

$= \sqrt{2} \sqrt{{tan}^{2} θ + 1} - ln (∣ ∣ \sqrt{{tan}^{2} θ + 1} + tan θ ∣ ∣)$ $=sqrt2sqrt(tan^2theta+1)-ln(abs(sqrt(tan^2theta+1)+tantheta))$

Using $tan θ = \frac{\sqrt{x} + 1}{\sqrt{2}}$ $tantheta=(sqrtx+1)/sqrt2$ :

$= \sqrt{2} \sqrt{\frac{{(\sqrt{x} + 1)}^{2}}{2} + 1} - ln ⎛ ⎜ ⎝ ∣ ∣ ∣ ∣ \sqrt{\frac{{(\sqrt{x} + 1)}^{2}}{2} + 1} + \frac{\sqrt{x} + 1}{\sqrt{2}} ∣ ∣ ∣ ∣ ⎞ ⎟ ⎠$ $=sqrt2sqrt((sqrtx+1)^2/2+1)-ln(abs(sqrt((sqrtx+1)^2/2+1)+(sqrtx+1)/sqrt2))$

$= \sqrt{2} \sqrt{\frac{{(\sqrt{x} + 1)}^{2} + 2}{2}} - ln ⎛ ⎜ ⎝ ∣ ∣ ∣ ∣ \sqrt{\frac{{(\sqrt{x} + 1)}^{2} + 2}{2}} + \frac{\sqrt{x} + 1}{\sqrt{2}} ∣ ∣ ∣ ∣ ⎞ ⎟ ⎠$ $=sqrt2sqrt(((sqrtx+1)^2+2)/2)-ln(abs(sqrt(((sqrtx+1)^2+2)/2)+(sqrtx+1)/sqrt2))$

$= \sqrt{{(\sqrt{x} + 1)}^{2} + 2} - ln ⎛ ⎜ ⎜ ⎝ ∣ ∣ ∣ ∣ ∣ \frac{\sqrt{{(\sqrt{x} + 1)}^{2} + 2} + \sqrt{x} + 1}{\sqrt{2}} ∣ ∣ ∣ ∣ ∣ ⎞ ⎟ ⎟ ⎠$ $=sqrt((sqrtx+1)^2+2)-ln(abs((sqrt((sqrtx+1)^2+2)+sqrtx+1)/sqrt2))$

Note that the $\frac{1}{\sqrt{2}}$ $1/sqrt2$ can actually be taken from the denominator of the logarithm as $- ln \sqrt{2}$ $-lnsqrt2$ , which will be absorbed into the constant of integration in the next step:

$= \sqrt{x + 2 \sqrt{x} + 3} - ln (∣ ∣ \sqrt{x} + \sqrt{x + 2 \sqrt{x} + 3} + 1 ∣ ∣) + C$ $=sqrt(x+2sqrtx+3)-ln(abs(sqrtx+sqrt(x+2sqrtx+3)+1))+C$

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How do you integrate $\int \frac{1}{\sqrt{4 x + 8 \sqrt{x} + 12}}$ $int 1/sqrt(4x+8sqrtx+12)$ using trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{4 x + 8 \sqrt{x} + 12}}$ $int 1/sqrt(4x+8sqrtx+12)$ using trigonometric substitution?