How do you integrate $\int \frac{1}{{(4 x^{2} - 9)}^{\frac{3}{2}}}$ $int 1/(4x^2-9)^(3/2)$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{{(4 x^{2} - 9)}^{\frac{3}{2}}}$ $int 1/(4x^2-9)^(3/2)$ by trigonometric substitution?

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Answer 1 · 2016-11-26T14:27:15

$\frac{- x}{9 \sqrt{4 x^{2} - 9}} + C$ $(-x)/(9sqrt(4x^2-9))+C$

Explanation:

$\int \frac{d x}{{(4 x^{2} - 9)}^{\frac{3}{2}}}$ $intdx/(4x^2-9)^(3/2)$

Let $2 x = 3 sec θ$ $2x=3sectheta$ such that $2 d x = 3 sec θ tan θ d θ$ $2dx=3secthetatanthetad theta$ .

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

$= \frac{1}{2} \int \frac{3 sec θ tan θ d θ}{{(9 {sec}^{2} θ - 9)}^{\frac{3}{2}}}$ $=1/2int(3secthetatanthetad theta)/(9sec^2theta-9)^(3/2)$

Factoring $9^{\frac{3}{2}} = 27$ $9^(3/2)=27$ from the denominator:

$= \frac{1}{2} \int \frac{3 sec θ tan θ}{27 {({sec}^{2} θ - 1)}^{\frac{3}{2}}} d θ$ $=1/2int(3secthetatantheta)/(27(sec^2theta-1)^(3/2))d theta$

Since ${sec}^{2} θ - 1 = {tan}^{2} θ$ $sec^2theta-1=tan^2theta$ :

$= \frac{1}{18} \int \frac{sec θ tan θ}{{tan}^{3} θ} d θ$ $=1/18int(secthetatantheta)/tan^3thetad theta$

$= \frac{1}{18} \int \frac{sec θ}{{tan}^{2} θ} d θ$ $=1/18intsectheta/tan^2thetad theta$

$= \frac{1}{18} \int \frac{cos θ}{{sin}^{2} θ} d θ$ $=1/18intcostheta/sin^2thetad theta$

Letting $u = sin θ$ $u=sintheta$ so $d u = cos θ d θ$ $du=costhetad theta$ .

$= \frac{1}{18} \int u^{- 2} d u$ $=1/18intu^-2du$

$= - \frac{1}{18 u}$ $=-1/(18u)$

$= - \frac{1}{18} csc θ$ $=-1/18csctheta$

$= - \frac{1}{18} \frac{sec θ}{tan θ}$ $=-1/18sectheta/tantheta$

$= - \frac{1}{54} \frac{3 sec θ}{\sqrt{{sec}^{2} θ - 1}}$ $=-1/54(3sectheta)/sqrt(sec^2theta-1)$

$= - \frac{3}{54} \frac{3 sec θ}{\sqrt{9 {sec}^{2} θ - 9}}$ $=-3/54(3sectheta)/sqrt(9sec^2theta-9)$

Using $3 sec θ = 2 x$ $3sectheta=2x$ :

$= - \frac{1}{18} \frac{2 x}{\sqrt{4 x^{2} - 9}}$ $=-1/18(2x)/sqrt(4x^2-9)$

$= \frac{- x}{9 \sqrt{4 x^{2} - 9}} + C$ $=(-x)/(9sqrt(4x^2-9))+C$

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How do you integrate $\int \frac{1}{{(4 x^{2} - 9)}^{\frac{3}{2}}}$ $int 1/(4x^2-9)^(3/2)$ by trigonometric substitution?

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