Evaluate the integral #int \ 1/(x^2sqrt(x^2-9)) \ dx #?

Question

Aurora Alvarado · Accepted Answer

# sqrt(x^2-9)/(9|x|)+C.#

We substitute, #x=3secy rArr dx=3secytanydy.#

#:. I=int1/(x^2sqrt(x^2-9))dx,#

#=int{(3secy tany)dy}/{9sec^2ysqrt(9sec^2y-9),#

#=int{(cancel(3)secycancel(tany))dy}/{9sec^2y*cancel(3tany)},#

#=1/9int1/secydy=1/9intcosydy,#

#=1/9siny,#

#=1/9sqrt(1-cos^2y),#

#=1/9sqrt(1-1/sec^2y),#

#=1/9sqrt(1-(3/x)^2).......................[because, x=3secy],#

#=1/9sqrt(x^2-9)/|x|.#

# rArr I=sqrt(x^2-9)/(9|x|)+C.#

Enjoy Maths.!

Paisley Johnson · Answer

# int \ 1/(x^2sqrt(x^2-9)) \ dx = 1/9 sqrt(1-9/x^2) + C #

We seek: # I = int \ 1/(x^2sqrt(x^2-9)) \ dx # We can write the integral as follows: # I = int \ 1/(x^2sqrt(9(x^2/9-1))) \ dx # # \ \ = int \ 1/(x^2sqrt(9)sqrt((x/3)^2-1)) \ dx # # \ \ = 1/3 \ int \ 1/(x^2 sqrt((x/3)^2-1)) \ dx # Let us attempt a substitution of the form: # 3sec theta = x => 3sectheta tantheta (d theta)/dx = 1 # Then substituting into the integral, we get: # I = 1/3 \ int \ 1/((3sec theta)^2 sqrt((sec theta)^2-1)) \ 3sectheta tantheta \ d theta # # \ \ = int \ (sectheta tantheta)/(9sec^2 theta sqrt(sec^2theta-1)) \ d theta # # \ \ = 1/9 \ int \ (tantheta)/(sec theta sqrt(tan^2 theta)) \ d theta # # \ \ = 1/9 \ int \ (tantheta)/(sec theta tan theta) \ d theta # # \ \ = 1/9 \ int \ (1)/(sec theta) \ d theta # # \ \ = 1/9 \ int \ cos theta \ d theta # # \ \ = 1/9 sin theta + C # # \ \ = 1/9 sqrt(1-cos^2 theta) + C # # \ \ = 1/9 sqrt(1-(1/sec^2 theta)) + C # And we can now restore the earlier substitution: # I = 1/9 sqrt(1-(1/(x/3)^2)) + C # # \ \ = 1/9 sqrt(1-(1/(x^2/9)) + C # # \ \ = 1/9 sqrt(1-9/x^2) + C #

Emilia Aguilar · Answer

#int dx/(x^2sqrt(x^2-9)) = sqrt(x^2-9)/(9x)+ C#

Restricting to the interval #x in (3,+oo)#, substitute:

#x=3sect#

#dx = 3 sect tant dt#

with #t in (0,pi/2)# and get:

#int dx/(x^2sqrt(x^2-9)) = 3 int (sect tant dt)/(9sec^2t sqrt(9sec^2t -9))#

#int dx/(x^2sqrt(x^2-9)) = 1/9 int ( tant dt)/(sect sqrt(sec^2t -1))#

use now the trigonometric identity:

#sec^2t -1 = 1/cos^2-1 = (1-cos^2t)/cos^2t =sin^2t/cos^2t = tan^2t#

so that:

#int dx/(x^2sqrt(x^2-9)) = 1/9 int ( tant dt)/(sect sqrt(tan^2t))#

and as in the selected interval the tangent is positive:

#int dx/(x^2sqrt(x^2-9)) = 1/9 int ( tant dt)/(sect tant) #

#int dx/(x^2sqrt(x^2-9)) = 1/9 int dt/sect #

#int dx/(x^2sqrt(x^2-9)) = 1/9 int costdt#

#int dx/(x^2sqrt(x^2-9)) = 1/9 sint + C#

To undo the substitution note that:

#x= 3sect = 3/cost = 3/sqrt(1-sin^2t)#

#sqrt(1-sin^2t) = 3/x#

#1-sin^2t = 9/x^2#

#sin^2t = 1-9/x^2#

and as in the interval the sine is positive:

#sint = sqrt(x^2-9)/x#

So:

#int dx/(x^2sqrt(x^2-9)) = sqrt(x^2-9)/(9x)+ C#

And using the substitution #x=-sect# for #x in (-oo,-3)# we obtain the same result.

Luke Alvarez · Answer

undefined To evaluate the integral $\int \frac{1}{x^2\sqrt{x^2-9}} \, dx$, we can use a trigonometric substitution. Recognizing that the integrand is of a form that suggests using the substitution $x = \sec(\theta)$ for $x^2 - 9$, because $x^2 = \sec^2(\theta)$ implies $x^2 - 9 = \tan^2(\theta)$, we proceed as follows:

1. **Substitution**: $x = \sec(\theta)$ implies $dx = \sec(\theta)\tan(\theta) d\theta$.
2. **Transform the Integral**: Substitute $x$ and $dx$ into the integral:

$$
\int \frac{1}{\sec^2(\theta)\sqrt{\tan^2(\theta)}} \sec(\theta)\tan(\theta) d\theta
$$

Since $\sec^2(\theta) = 1 + \tan^2(\theta)$, and given our transformation, the integral simplifies to:

$$
\int \frac{\sec(\theta)\tan(\theta)}{\sec^2(\theta)\tan(\theta)} d\theta = \int \frac{1}{\sec(\theta)} d\theta = \int \cos(\theta) d\theta
$$

3. **Integrate**: Integrating $\cos(\theta)$ gives:

$$
\sin(\theta) + C
$$

4. **Back-substitute**: Recall that $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ in the context of our substitution $x = \sec(\theta) = \frac{1}{\cos(\theta)}$, which implies $x = \frac{\text{hypotenuse}}{\text{adjacent}}$, and therefore, using the Pythagorean identity, we find that the opposite side (for $\sin(\theta)$) is $\sqrt{x^2 - 9}$. So, $\sin(\theta) = \frac{\sqrt{x^2 - 9}}{x}$.

Thus, the final answer is:

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

This is the evaluated form of the integral $\int \frac{1}{x^2\sqrt{x^2-9}} \, dx$.