How do you evaluate $\int$ $int$ $\frac{d x}{x^{2} \sqrt{x^{2} - 9}}$ $dx/(x^2sqrt(x^2 - 9))$ with x = 3sec( $θ$ $theta$ )?

Question

Evaluate the integral using the indicated trigonometric substitution. (Use C for the constant of integration.)

$\int$ $int$ $\frac{d x}{x^{2} \sqrt{x^{2} - 9}}$ $dx/(x^2sqrt(x^2 - 9))$ , x = 3sec( $θ$ $theta$ )

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Answer 1 · 2018-07-05T11:29:58

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

Let $x = 3 sec θ$ $x = 3sectheta$

$\Rightarrow d x = 3 sec θ tan θ d θ$ $=>dx = 3sectheta tantheta d theta$

So
$I = \int \frac{3 sec θ tan θ d θ}{9 {sec}^{2} θ \sqrt{9 {sec}^{2} θ - 9}}$ $I=int(3sectheta tantheta d theta)/(9sec^2thetasqrt(9sec^2theta- 9))$

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

$= \frac{1}{9} \int \frac{sec θ tan θ d θ}{{sec}^{2} θ tan θ}$ $=1/9int(sectheta tantheta d theta)/(sec^2theta tan theta)$

$= \frac{1}{9} \int cos θ d θ$ $=1/9intcosthetad theta$

$= \frac{1}{9} sin θ + C$ $=1/9sintheta +C$

$= \frac{1}{9} \frac{tan θ}{sec θ} + C$ $=1/9tantheta/sectheta +C$

$= \frac{1}{9} \frac{\sqrt{{sec}^{2} θ - 1}}{sec θ} + C$ $=1/9(sqrt(sec^2theta-1))/sectheta +C$

$= \frac{1}{9} \frac{\sqrt{\frac{x^{2}}{9} - 1}}{\frac{x}{3}} + C$ $=1/9(sqrt(x^2/9-1))/(x/3) +C$

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