Integration Help for \int 1/sqrt(a^2 + x^2)

In summary, the given problem involves finding the integral of 1 divided by the square root of a squared plus x squared. After some attempts, the solution is found to be equal to the square root of c times the integral of 1 divided by the square root of c times x squared plus 1. This is equivalent to c times the inverse sine of c times x, divided by c. However, this is not the final solution as it is not an arcsine integral. The correct solution involves substituting x with a tangent of theta.
  • #1
holezch
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Homework Statement




[tex]\int \frac{1}{\sqrt{a^{2} + x^{2}}}[/tex]

Homework Equations


The Attempt at a Solution



I got:[tex]\int \frac{1}{\sqrt{a^{2} + x^{2}}} = \sqrt{c} \ast \int \frac{1}{\sqrt{c \ast{x^{2}} + 1}} [/tex] [tex] \frac{1}{a^{2}} = c [/tex]

NEVERMIND! I got it: I couldn't remember the integral of arcsin..
so it's [tex] \frac{1}{a} = c [/tex][tex] c \ast \int \frac{1}{\sqrt{c \ast{x^{2}} + 1}}

= c \ast arcsin(cx) /c [/tex]
 
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  • #2
holezch said:

Homework Statement




[tex]\int \frac{1}{\sqrt{a^{2} + x^{2}}}[/tex]


Homework Equations





The Attempt at a Solution



I got:


[tex]\int \frac{1}{\sqrt{a^{2} + x^{2}}} = \sqrt{c} \ast \int \frac{1}{\sqrt{c \ast{x^{2}} + 1}} [/tex]


[tex] \frac{1}{a^{2}} = c [/tex]

NEVERMIND! I got it: I couldn't remember the integral of arcsin..
so it's


[tex] \frac{1}{a} = c [/tex]


[tex] c \ast \int \frac{1}{\sqrt{c \ast{x^{2}} + 1}}

= c \ast arcsin(cx) /c [/tex]


This is not an arcsine integral. :redface:
Try [tex]x=a \,\ tan(\theta)[/tex].
 

FAQ: Integration Help for \int 1/sqrt(a^2 + x^2)

What is the general form of the integral \int \frac{1}{\sqrt{a^2 + x^2}}?

The general form of the integral \int \frac{1}{\sqrt{a^2 + x^2}} is \ln|a + \sqrt{a^2 + x^2}| + C, where C is a constant of integration.

How is the integral \int \frac{1}{\sqrt{a^2 + x^2}} evaluated?

The integral \int \frac{1}{\sqrt{a^2 + x^2}} is evaluated using the trigonometric substitution x = a\tan\theta, which transforms the integral into \int \frac{1}{a^2 + a^2\tan^2\theta}\sec^2\theta d\theta. This can then be solved using basic trigonometric identities.

What is the domain of the function \frac{1}{\sqrt{a^2 + x^2}}?

The domain of the function \frac{1}{\sqrt{a^2 + x^2}} is all real numbers except for x = \pm a, as this would result in a division by zero.

Can the integral \int \frac{1}{\sqrt{a^2 + x^2}} be solved using other methods?

Yes, the integral \int \frac{1}{\sqrt{a^2 + x^2}} can also be solved using the u-substitution x = a\tan u, or by using a trigonometric table to directly evaluate the integral.

What is the significance of the constant a in the integral \int \frac{1}{\sqrt{a^2 + x^2}}?

The constant a represents the distance of the point (x, \frac{1}{\sqrt{a^2 + x^2}}) from the origin on the x-axis. This integral is commonly used in physics to calculate the electric potential due to a point charge located at the origin.

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