Trig substitution should be simple but it's driving me nuts

In summary, the conversation discusses how to evaluate the integral \int \frac{x}{\sqrt{x^2+4}}\,dx using substitution. The correct substitution to use is \theta = \tan^{-1}(x/2), which simplifies the integral to \int \frac{2 \tan \theta \ \sec^2\theta}{\sqrt{4\tan^2\theta+4}}\,d\theta. However, the integral is still incorrect due to missing the dx term. After correcting this mistake, the conversation also points out another simpler substitution and thanks the others for their help.
  • #1
pugtm
18
0

Homework Statement


[itex]\int x/sqrt{(x^{2}+4)}[/itex]


Homework Equations


x=2tanx


The Attempt at a Solution


x=2tanx
[itex]\int[/itex]2tan[itex]\vartheta[/itex]/[itex]\sqrt{tan^2\vartheta}[/itex]+4
2/2 *[itex]\int[/itex]tan/sec

[itex]\int[/itex]sin=-cos

now is the part where i am stuck
i know from using substitution that the answr should be [itex]\sqrt{x^2+4}[/itex]
but no matter how i manipulate it, it comes out strange.
all help is appreciated
 
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  • #2
The integral you're trying to evaluate is
[tex]\int\frac{x}{\sqrt{x^2+4}}\,dx[/tex]
Note the presence of the dx. That's what you forgot to account for when you did the trig substitution.
 
  • #3
how do you get rid of the natural log at the end?
 
  • #4
What natural log? You apparently made another mistake.
 
  • #5
the new integral is
int 2tan(x)sec(x)^2/Sqrt(4tan(x)^2+4)
the sec^2 cancel out making it the integral of 2tan(x)=-2lncos(x)
 
  • #6
Slow down and look at it more carefully.
[tex]\int \frac{2 \tan \theta \ \sec^2\theta}{\sqrt{4\tan^2\theta+4}}\,d\theta = \int \frac{2 \tan \theta \ \sec^2\theta}{2\sqrt{\sec^2\theta}}\,d\theta[/tex]
 
  • #7
pugtm said:
i know from using substitution that the answr should be [itex]\sqrt{x^2+4}[/itex]
but no matter how i manipulate it, it comes out strange.
all help is appreciated

I'm not sure that's what you mean, but I hope you noticed that there's a much simpler substitution possible :-)

Anyway, the others are correct about your mistakes :-)
 
  • #8
thank you all for your timely assistance
 

1. Why is trig substitution necessary in math?

Trig substitution is a technique used to solve integrals involving trigonometric functions. It allows us to transform an integral into a form that is easier to solve using basic trigonometric identities.

2. How do I know when to use trig substitution?

Trig substitution is used when the integral involves a square root of a quadratic expression or an expression containing a sum or difference of squares. It is also useful when the integral involves trigonometric functions raised to an odd power.

3. What are the steps for using trig substitution?

The first step is to identify the type of substitution needed based on the form of the integral. Then, choose an appropriate substitution by setting one of the trigonometric functions equal to the variable in the integral. After substitution, simplify the integral using trigonometric identities and solve for the variable. Finally, substitute back in the original variable to get the final answer.

4. Why is trig substitution considered difficult?

Trig substitution can be challenging because it requires a good understanding of trigonometric identities and the ability to recognize which substitution to use for a given integral. It also involves several steps and can be time-consuming.

5. Are there any tips for mastering trig substitution?

Practice is key when it comes to mastering trig substitution. Familiarize yourself with common trigonometric identities and practice solving a variety of integrals using trig substitution. It also helps to break down the process into smaller steps and work through each one carefully.

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