Integrating: \int{\frac{xsec^2x}{tanx+\sqrt{3}}}dx

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This one can be integrated using integration by parts.u = tan-1x, dv = dxdu = dx/(1 + x2), v = xThis results in\int tan^{-1}x dx = xtan^{-1}x - \int xdx/(1 + x^2)The latter integral can be done with an ordinary substitution, u = 1 + x2.In summary, to solve the integral \int{\frac{xsec^2x}{tanx+\sqrt{3}}}dx, the initial substitution of u = tanx + sqrt{3} and dx = du/sec^2x can be made. However, the integral cannot be solved using elementary functions and alternative methods such as
  • #1
Mentallic
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Homework Statement



[tex]\int{\frac{xsec^2x}{tanx+\sqrt{3}}}dx[/tex]

The Attempt at a Solution



I let [tex]tanx+\sqrt{3}=u[/tex]

Then, [tex]\frac{du}{dx}=sec^2x , dx=\frac{du}{sec^2x}[/tex]

Substituting into the integral: [tex]\int{\frac{x}{u}du[/tex]

But now I don't know what to do. That little x is really screwing me over on this... Maybe a whole other approach needs to be taken?
 
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  • #2
From your substitution it follows that [itex]x=\arctan(u-\sqrt{3})[/itex]. That said I don't think this integral has a solution within the elementary functions. It will involve some kind of poly logarithm.
 
  • #3
If this integral has no elementary solution, I think I may give up and walk away from it.

Just out of curiosity, what would [tex]\int{arctanx}dx[/tex] be? I'm sure it's not elementary but the anti-derivative must've been found/solved for such a simple result?
 
  • #4
Mentallic said:
If this integral has no elementary solution, I think I may give up and walk away from it.

Just out of curiosity, what would [tex]\int{arctanx}dx[/tex] be? I'm sure it's not elementary but the anti-derivative must've been found/solved for such a simple result?
This one can be integrated using integration by parts.
u = tan-1x, dv = dx
du = dx/(1 + x2), v = x

This results in
[tex]\int tan^{-1}x dx = xtan^{-1}x - \int xdx/(1 + x^2)[/tex]
The latter integral can be done with an ordinary substitution, u = 1 + x2.
 
  • #5
That one is simpler than you think, write it as [itex] \int 1*\arctan x dx[/itex], then use partial integration.
 
  • #6
I think I might start reading up on integration by parts, since I couldn't follow where the v came from and the dv=dx, v=x?

If it comes down to not being able to solve an integral with elementary functions, are there any methods available to solve one as such? Other than taking guesses and differentiating to be sure :smile:
 
  • #7
Mentallic said:
I think I might start reading up on integration by parts, since I couldn't follow where the v came from and the dv=dx, v=x?
That's exactly where the v came from: if dv= dx, then, integrating both sides, v= x!

If it comes down to not being able to solve an integral with elementary functions, are there any methods available to solve one as such? Other than taking guesses and differentiating to be sure :smile:
If it really is true that a given integral cannot be done with "elementary functions", then what in the world would you "guess"? It not a matter of certain methods not working, it is that the answer itself cannot be written in terms of elementary functions. In that case you either define a new function to be its integral or try to convert to a such an integral for which the integral has already been defined.

For example,
[tex]\int e^{-x^2}dx[/tex]
cannot be done "in terms of elementary functions" so the non-elementary function Erf(x) (the "error" function because that integral shows up in probability and calculating "random errors") is defined as that integral.
[tex]\int e^{-x^2}dx= Erf(x)+ C[/tex]

But if you had, now
[tex]\int e^{-(2x-3)^2} dx[/tex]
you could make the substitution u= 2x- 3 so that du= 2dx and dx= (1/2)du so
[tex]\int e^{-(2x-3)^2} dx= \frac{1}{2}\int e^{-u^2}du= \frac{1}{2}Erf(u)+ C= \frac{1}{2}Erf(2x-3)+ C[/tex].
 
  • #8
HallsofIvy said:
That's exactly where the v came from: if dv= dx, then, integrating both sides, v= x!
Not sure what the v is doing there in the first place. Oh thanks, I could've guessed that if dv=dx then v=x but I never quite trusted any algebraic manipulations of dx.


HallsofIvy said:
If it really is true that a given integral cannot be done with "elementary functions", then what in the world would you "guess"? It not a matter of certain methods not working, it is that the answer itself cannot be written in terms of elementary functions.
Ahh then I was misunderstanding 'elementary functions'. I suppose I confused it with elementary methods of integrating.
This is quite interesting! Has it been proven that [tex]\int{e^{-x^2}}dx[/tex] cannot be expressed with elementary functions, or is it just that the integral (still possible to express with elementary functions) has yet to be found?
 
  • #9
It has been proven that it does not exist. As for integration by parts, if you find differentials tricky you can also write it like this.

[tex]\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x) dx[/tex].

Now write [itex]\int \arctan x dx =\int 1*\arctan x dx[/itex]. Take [itex]f(x)=\arctan x, \;\;\; g'(x)=1[/itex]. Now try to apply integration by parts.
 
  • #10
Integration by parts is the integration counterpart to the product rule of differentiation.
d/dx(f(x)*g(x)) = df(x)/dx*g(x) + f(x)*dg(x)/dx
Multiply by dx to get d(f(x) * g(x)) = df(x) * g(x) + f(x)*dg(x)

Antidifferentiate the above to get
[itex]\int d(f(x) * g(x)) = \int df(x) * g(x) + \int f(x)*dg(x)[/itex]

or
[itex]f(x)*g(x) = \int df(x) * g(x) + \int f(x)*dg(x)[/itex]
or finally,
[itex]\int df(x) * g(x) = f(x)*g(x) - \int f(x)*dg(x)[/itex]
 

FAQ: Integrating: \int{\frac{xsec^2x}{tanx+\sqrt{3}}}dx

What is the process for integrating this expression?

To integrate this expression, you can use the substitution method. Let u = tanx + √3, du = sec^2x dx. Substituting these values into the integral gives ∫(x/u) du, which can be easily integrated using the power rule.

What are the limits of integration for this expression?

The limits of integration will depend on the specific application of this integral. Generally, if the integral represents the area under the curve of the given function, the limits will correspond to the x-values where the curve intersects the x-axis.

Can this integral be evaluated using other methods?

Yes, this integral can also be evaluated using the u-substitution method or by using trigonometric identities to simplify the expression. However, the substitution method is the most straightforward approach for this particular expression.

What are the possible applications of this integral?

Integrals involving trigonometric functions are commonly used in physics, engineering, and other scientific fields. This particular integral may be useful in calculating the work done by a force applied at an angle, or in calculating the power output of a rotating object.

Is there a way to check the accuracy of the integrated result?

Yes, you can check the accuracy of the integrated result by differentiating it. If the derivative of the integrated expression matches the original integrand, then the result is accurate. Additionally, you can also use numerical methods to approximate the integral and compare it to the integrated result.

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