Question #34e5a

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Question #34e5a

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Answer 1 · 2017-09-01T19:56:49

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

Explanation:

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

Use the substitution $x = 3 sec θ$ $x=3sectheta$ . This implies that $d x = 3 sec θ tan θ d θ$ $dx=3secthetatanthetad theta$ . Importantly, $x^{2} - 9 = 9 {sec}^{2} θ - 9 = 9 ({sec}^{2} θ - 1) = 9 {tan}^{2} θ$ $x^2-9=9sec^2theta-9=9(sec^2theta-1)=9tan^2theta$ .

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

$I = \int \frac{27 {sec}^{3} θ tan θ}{27 {tan}^{3} θ} d θ$ $I=int(27sec^3thetatantheta)/(27tan^3theta)d theta$

$I = \int \frac{{sec}^{3} θ}{{tan}^{2} θ} d θ$ $I=intsec^3theta/tan^2thetad theta$

$I = \int \frac{1}{{cos}^{3} θ} \frac{{cos}^{2} θ}{{sin}^{2} θ}$ $I=int1/cos^3thetacos^2theta/sin^2theta$

$I = \int {csc}^{2} θ sec θ d θ$ $I=intcsc^2thetasecthetad theta$

Use ${csc}^{2} θ = {cot}^{2} θ + 1$ $csc^2theta=cot^2theta+1$ :

$I = \int ({cot}^{2} θ + 1) sec θ d θ$ $I=int(cot^2theta+1)secthetad theta$

$I = \int (\frac{{cos}^{2} θ}{{sin}^{2} θ} \frac{1}{cos θ} + sec θ) d θ$ $I=int(cos^2theta/sin^2theta 1/costheta+sectheta)d theta$

$I = \int (cot θ csc θ + sec θ) d θ$ $I=int(cotthetacsctheta+sectheta)d theta$

These have common integrals:

$I = - csc θ + ln | sec θ + tan θ | + C$ $I=-csctheta+lnabs(sectheta+tantheta)+C$

Our original substitution was $sec θ = \frac{x}{3}$ $sectheta=x/3$ . This means that $cos θ = \frac{3}{x}$ $costheta=3/x$ , so we have a right triangle where the side adjacent to $θ$ $theta$ is $3$ $3$ and the hypotenuse is $x$ $x$ . Through the Pythagorean Theorem, the side opposite $θ$ $theta$ is $\sqrt{x^{2} - 9}$ $sqrt(x^2-9)$ . Then:

$tan θ = \frac{opp}{adj} = \frac{\sqrt{x^{2} - 9}}{3}$ $tantheta="opp"/"adj"=sqrt(x^2-9)/3$
$csc θ = \frac{1}{sin θ} = \frac{hyp}{opp} = \frac{x}{\sqrt{x^{2} - 9}}$ $csctheta=1/sintheta="hyp"/"opp"=x/sqrt(x^2-9)$

Hence:

$I = ln ∣ ∣ ∣ \frac{x}{3} + \frac{\sqrt{x^{2} + 9}}{3} ∣ ∣ ∣ - \frac{x}{\sqrt{x^{2} - 9}} + C$ $I=lnabs(x/3+sqrt(x^2+9)/3)-x/sqrt(x^2-9)+C$

Factor $\frac{1}{3}$ $1/3$ from the natural logarithm. Using $log (A B) = log (A) + log (B)$ $log(AB)=log(A)+log(B)$ , a $ln (\frac{1}{3})$ $ln(1/3)$ will be spit out of the logarithm but it can be neglected because as a constant it will be absorbed into $C$ $C$ , the constant of integration.

$I = ln ∣ ∣ x + \sqrt{x^{2} - 9} ∣ ∣ - \frac{x}{\sqrt{x^{2} - 9}} + C$ $I=lnabs(x+sqrt(x^2-9))-x/sqrt(x^2-9)+C$

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Question #34e5a

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