Trig Substitution: Solve Int \sqrt{x^2 + 16}

[teneleven]

Homework Statement

[tex]\int\frac{dx}{\sqrt{x^2 + 16}}[/tex]

Homework Equations

[tex]x = 4\tan\theta[/tex]
[tex]dx = 4\sec^2\theta \ d\theta[/tex]

The Attempt at a Solution

[tex]\int\frac{4\sec^2\theta}{\sqrt{16\tan^2\theta + 16}}\ d\theta[/tex]

[tex]\int\frac{4\sec^2\theta}{\sqrt{16(\tan^2\theta + 1)}}\ d\theta[/tex]

[tex]\int\frac{4\sec^2\theta}{4\sec\theta}\ d\theta[/tex]

[tex]\int\sec\theta\ d\theta[/tex]

How do I reduce past this step?
Integration by parts returns me to [tex]\int\sec\theta\ d\theta[/tex]

The answer at the back of the book is as follows: [tex]\ln(\sqrt{x^2 + 16} + x) + C[/tex]

Thanks.

EDIT: notational mistakes corrected.

 

[neutrino]

Nothing to see here...move on.

 

[ChaoticLlama]

Isn't that supposed to be [tex]16\int{\sec^3{\theta}}[/tex]?

Nope, his integral is correct.

And the answer in the book is also correct.

The integral of secant is not very trivial, and the best method is to multiply by '1'.
In this case by (sec@ + tan@)/(sec@ + tan@).
Then do a 'u' substitution if the integral isn't all that obvious yet.
(let u = sec@ + tan@)

enjoy!

P.S. Teneleven -> you notated your trig-sub wrong, yet used it correctly.
P.P.S. sec³@ is significantly harder it integrate the first time through.

 

[neutrino]

Nope, his integral is correct.

And the answer in the book is also correct.

Oh. Well, the first time I saw the problem the radical was in the numerator.

 

[Gib Z]

Just incase your wondering, by multiply by one, he's hinting at Integration by parts.

 

[HallsofIvy]

The way I would integrate sec x, not necessarily the simplest, is to write sec x as 1/cos x, multiply both numerator and denominator by cos x:
[tex]\int \frac{cos x dx}{cos^2 x}[/tex]
rewrite as
[tex]\int \frac{cos x dx}{1- sin^2 x}[/tex]
Let u= sin(x) and use partial fractions on the remaining integral.

 

[dextercioby]

Make [itex] x=4\sinh t [/itex] and see what you get. Don't forget the integration constant.

 

[teneleven]

I solved it using ChaoticLlama's method, but without integration by parts. I'm having trouble reducing it down to the answer in the book. I end up with a "4" in the denominator and haven't figured out how to get rid of it.

[tex]\int\sec\theta\frac{(\sec\theta + \tan\theta)}{(\sec\theta + \tan\theta)}\ d\theta[/tex]

[tex]\int\frac{\sec^2\theta}{\sec\theta + \tan\theta}\ d\theta + \int\frac{\sec\theta\tan\theta}{\sec\theta +\tan\theta}\ d\theta[/tex]

[tex]\ln(\sec\theta + \tan\theta) + C[/tex]

[tex]\ln(\frac{\sqrt{x^2 + 16}}{4} + \frac{x}{4}) + C[/tex]

 

[ChaoticLlama]

You've done exactly as required.

Just Remember the property of logarithms ln(a/b) = ln(a) - ln(b)

Then you'll recognize that...
[tex]\ln(\frac{\sqrt{x^2 + 16}}{4} + \frac{x}{4}) + C =
\ln(\sqrt{x^2 + 16} + x) + D[/tex]

(What does D equal with respect to C?)

 

[teneleven]

Thanks for the prompt reply.

I'm looking at it and I recall the property of logarithms, but I still don't understand how the "4" is reduced to [tex]\ln(4)[/tex] and removed in the final answer.

EDIT: So you're saying that D = C + [tex]\ln(4)[/tex]?

 

[Gib Z]

Yes that's exactly what he's saying. Remember C is just any constant, when you differentiate it disappears! ln4 is also just a constant, A constant plus another constant is just another constant!

BTW Chaoticlama, No Integration by parts for sec x will not help in the least, Halls of Ivys method got me the answer fine though.