How do you integrate $\int \sqrt{e^{8 x} - 9}$ $int sqrt(e^(8x)-9)$ using trig substitutions?

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How do you integrate $\int \sqrt{e^{8 x} - 9}$ $int sqrt(e^(8x)-9)$ using trig substitutions?

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Answer 1 · 2016-09-15T23:43:39

$\frac{\sqrt{e^{8 x} - 9} - 3 arcsec (\frac{e^{4 x}}{3})}{4} + C$ $(sqrt(e^(8x)-9)-3"arcsec"(e^(4x)/3))/4+C$

Explanation:

$\int \sqrt{e^{8 x} - 9} d x$ $intsqrt(e^(8x)-9)dx$

Apply the substitution $e^{4 x} = 3 sec θ$ $e^(4x)=3sectheta$ . Thus $4 e^{4 x} d x = 3 sec θ tan θ d θ$ $4e^(4x)dx=3secthetatanthetad theta$ .

Note that $d x = \frac{3 sec θ tan θ d θ}{4 e^{4 x}} = \frac{3 sec θ tan θ d θ}{4 (3 sec θ)} = \frac{1}{4} tan θ d θ$ $dx=(3secthetatanthetad theta)/(4e^(4x))=(3secthetatanthetad theta)/(4(3sectheta))=1/4tanthetad theta$ .

$= \int \sqrt{{(e^{4 x})}^{2} - 9} d x$ $=intsqrt((e^(4x))^2-9)dx$

$= \int \sqrt{9 {sec}^{2} θ - 9} (\frac{1}{4} tan θ d θ)$ $=intsqrt(9sec^2theta-9)(1/4tanthetad theta)$

$= \frac{3}{4} \int \sqrt{{sec}^{2} θ - 1} (tan θ d θ)$ $=3/4intsqrt(sec^2theta-1)(tanthetad theta)$

Note that ${tan}^{2} θ = {sec}^{2} θ - 1$ $tan^2theta=sec^2theta-1$ , so:

$= \frac{3}{4} \int {tan}^{2} θ d θ$ $=3/4inttan^2thetad theta$

$= \frac{3}{4} \int ({sec}^{2} θ - 1) d θ$ $=3/4int(sec^2theta-1)d theta$

$= \frac{3}{4} \int {sec}^{2} θ d θ - \frac{3}{4} \int d θ$ $=3/4intsec^2thetad theta-3/4intd theta$

$= \frac{3}{4} tan θ - \frac{3}{4} θ + C$ $=3/4tantheta-3/4theta+C$

From $e^{4 x} = 3 sec θ$ $e^(4x)=3sectheta$ , we see that $θ = arcsec (\frac{e^{4 x}}{3})$ $theta="arcsec"(e^(4x)/3)$ . Also, since $sec θ = \frac{e^{4 x}}{3}$ $sectheta=e^(4x)/3$ , we see that the hypotenuse is $e^{4 x}$ $e^(4x)$ , the adjacent side is $3$ $3$ , and the opposite side is $\sqrt{e^{8 x} - 9}$ $sqrt(e^(8x)-9)$ . Thus $tan θ = \frac{\sqrt{e^{8 x} - 9}}{3}$ $tantheta=sqrt(e^(8x)-9)/3$ .

$= \frac{3}{4} (\frac{\sqrt{e^{8 x} - 9}}{3}) - \frac{3}{4} arcsec (\frac{e^{4 x}}{3}) + C$ $=3/4(sqrt(e^(8x)-9)/3)-3/4"arcsec"(e^(4x)/3)+C$

$= \frac{\sqrt{e^{8 x} - 9} - 3 arcsec (\frac{e^{4 x}}{3})}{4} + C$ $=(sqrt(e^(8x)-9)-3"arcsec"(e^(4x)/3))/4+C$

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How do you integrate $\int \sqrt{e^{8 x} - 9}$ $int sqrt(e^(8x)-9)$ using trig substitutions?

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How do you integrate $\int \sqrt{e^{8 x} - 9}$ $int sqrt(e^(8x)-9)$ using trig substitutions?