Trig Substitution for Integrating \frac{dx}{\sqrt{x^{2}+16}}

In summary, the integral of dx/sqrt(x^2+16) can be solved by substituting x=4tanθ and dx=4sec^2θdθ. This results in ln|sec(√(x^2+16)/4)+tan(x/4)| + C as the final answer, with C being a constant of integration. Hyperbolic functions can also be used to solve this integral, but it may not make much difference in this case.
  • #1
iRaid
559
8

Homework Statement


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

Homework Equations


The Attempt at a Solution


[tex]x=4tan\theta[/tex] [tex]dx=4sec^{2}\theta d\theta[/tex]
Therefore:
[tex]\int \frac{4sec^{2}\theta d\theta}{\sqrt{16tan^{2}\theta +16}} = \int \frac{sec^{2}\theta d\theta}{\sqrt{tan^{2}\theta+1}}[/tex]
[tex]\int \frac{sec^{2}\theta d\theta}{sec\theta} = \int sec\theta d\theta[/tex]
[tex]=ln|sec\theta+tan\theta| = ln|sec\frac{\sqrt{x^{2}+16}}{4}+tan\frac{x}{4}|[/tex]

Wolfram is getting a hyperbolic sine function so idk what is wrong (we've never talked about hyperbolic functions in class, so I don't think that's the answer)
 
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  • #2
You are OK at ##\ln|\sec\theta+\tan\theta|##. But in your very last expression you should leave out the secant and tangent, since you have calculated them in terms of x already. And you need to add a constant of integration. Then your answer will be correct.

Don't worry about the hyperbolic functions. Integrals like this can be worked either way.
 
  • #3
LCKurtz said:
You are OK at ##\ln|\sec\theta+\tan\theta|##. But in your very last expression you should leave out the secant and tangent, since you have calculated them in terms of x already. And you need to add a constant of integration. Then your answer will be correct.

Don't worry about the hyperbolic functions. Integrals like this can be worked either way.

Oh right (feel stupid lol). Are the hyperbolic functions an easier way to solve these problems?

Thanks for the help
 
  • #4
iRaid said:
Oh right (feel stupid lol). Are the hyperbolic functions an easier way to solve these problems?

Thanks for the help

They can be easier, but in this case I don't think it makes much difference. Try ##x = 4\sinh t## on it and see what you think. Remember the basic hyperbolic identity is ##\cosh^2t - \sinh^2 t = 1##.
 

1. What is trig substitution in integrals?

Trig substitution is a method used to simplify integrals that involve radicals and algebraic expressions by replacing them with trigonometric functions.

2. When should I use trig substitution in an integral?

Trig substitution is typically used when you have an integral with a radical expression that can be simplified using a trigonometric identity.

3. How do I choose which trigonometric function to substitute with?

The choice of trigonometric function depends on the form of the radical expression in the integral. For example, if the expression involves a square root of a quadratic equation, you can use the sine or cosine function. If it involves a square root of a sum or difference of squares, you can use the tangent function.

4. Can I use trig substitution for all types of integrals?

No, trig substitution is only applicable to specific types of integrals, particularly those involving radicals and algebraic expressions. It is not always the most efficient method and may not work for all integrals.

5. Are there any specific steps to follow when using trig substitution in an integral?

Yes, there are general steps to follow when using trig substitution. First, identify the appropriate substitution based on the form of the expression. Then, substitute the expression with the trigonometric function and simplify the integral. Finally, use trigonometric identities to solve the integral and substitute back the original variable to get the final answer.

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