Trig Substitution

teneleven

Homework Statement

$$\int\frac{dx}{\sqrt{x^2 + 16}}$$

Homework Equations

$$x = 4\tan\theta$$
$$dx = 4\sec^2\theta \ d\theta$$

The Attempt at a Solution

$$\int\frac{4\sec^2\theta}{\sqrt{16\tan^2\theta + 16}}\ d\theta$$

$$\int\frac{4\sec^2\theta}{\sqrt{16(\tan^2\theta + 1)}}\ d\theta$$

$$\int\frac{4\sec^2\theta}{4\sec\theta}\ d\theta$$

$$\int\sec\theta\ d\theta$$

How do I reduce past this step?
Integration by parts returns me to $$\int\sec\theta\ d\theta$$

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

Thanks.

EDIT: notational mistakes corrected.

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neutrino
Nothing to see here...move on.

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ChaoticLlama
Isn't that supposed to be $$16\int{\sec^3{\theta}}$$?

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.

Homework Helper
Just incase your wondering, by multiply by one, hes hinting at Integration by parts.

Homework Helper
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:
$$\int \frac{cos x dx}{cos^2 x}$$
rewrite as
$$\int \frac{cos x dx}{1- sin^2 x}$$
Let u= sin(x) and use partial fractions on the remaining integral.

Homework Helper
Make $x=4\sinh t$ 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.

$$\int\sec\theta\frac{(\sec\theta + \tan\theta)}{(\sec\theta + \tan\theta)}\ d\theta$$

$$\int\frac{\sec^2\theta}{\sec\theta + \tan\theta}\ d\theta + \int\frac{\sec\theta\tan\theta}{\sec\theta +\tan\theta}\ d\theta$$

$$\ln(\sec\theta + \tan\theta) + C$$

$$\ln(\frac{\sqrt{x^2 + 16}}{4} + \frac{x}{4}) + C$$

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...
$$\ln(\frac{\sqrt{x^2 + 16}}{4} + \frac{x}{4}) + C = \ln(\sqrt{x^2 + 16} + x) + D$$

(What does D equal with respect to C?)

teneleven
I'm looking at it and I recall the property of logarithms, but I still don't understand how the "4" is reduced to $$\ln(4)$$ and removed in the final answer.
EDIT: So you're saying that D = C + $$\ln(4)$$?