Are Both Answers Correct for Trigonometric Substitution Integral?

In summary: Thanks!I think I fully understand what is going on now and PeroK's formula, which is obvious when you see it, makes it even clearer. Thanks!
  • #1
Rhapsody83
6
3
Last night I tried to calculate from an automatically generated Wolfram Alpha problem set:
$$\int{\frac{1}{\sqrt{x^2+4}}}dx$$
I answered $$\ln({\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}})+C$$
The answer sheet gave:
$$\ln({\sqrt{x^2+4}+x})+C$$
I couldn't see where I had gone wrong, so I tried differentiating my answer in Wolfram Alpha and the answer sheet one as well and both gave
$$\frac{1}{\sqrt{x^2+4}}$$
I'm confused as this suggests that both are correct answers to the question. Can somebody please tell me what is going on as I thought every function had a unique integral? Are both or only one of the answers correct?

PS There should be absolute value brackets around my integral answers but I'm not sure how to do them in LaTeX.
 
  • Like
Likes PeroK
Physics news on Phys.org
  • #2
You didn't make a mistake. The two expressions are the same with only different constants.
$$
\log\left(\dfrac{\sqrt{x^2+4}}{2}+\dfrac{x}{2}\right)+C=\log\left(\dfrac{1}{2}\cdot\left(\sqrt{x^2+4}+x\right)\right)+C=\log\left(\sqrt{x^2+4}+x\right)+\underbrace{\log\left(\dfrac{1}{2}\right)+C}_{=:C'}
$$
 
  • Like
  • Informative
Likes vanhees71, jedishrfu, PeroK and 1 other person
  • #3
Thanks for this, I was very confused, but it makes sense now. Am I right in thinking that one that is not divided by two is the preferred one as it is a little simpler?
 
  • #4
Rhapsody83 said:
Thanks for this, I was very confused, but it makes sense now. Am I right in thinking that one that is not divided by two is the preferred one as it is a little simpler?
It is finally a matter of taste. I would prefer the version without an additional factor that is swallowed by the constant anyway. I mean, you could even write ##\log\left(\dfrac{\sqrt{x^2+4}}{\pi}+ \dfrac{x}{\pi}\right),## which, of course, doesn't make much sense. So, yes, put all unnecessary factors in the logarithm (summands outside the logarithm) into ##C.##

This question doesn't come up if you consider definite integrals.
 
  • Like
  • Informative
Likes vanhees71 and Rhapsody83
  • #5
... more generally:
$$\frac d {dx}\bigg ( \ln (\frac{f(x)}{k}) \bigg ) = \frac d {dx}\bigg ( \ln (f(x)) - \ln k \bigg ) = \frac d {dx}\bigg ( \ln (f(x))\bigg )$$Where ##k## is a constant.
 
  • Like
  • Informative
Likes vanhees71 and Rhapsody83
  • #6
I think I fully understand what is going on now and PeroK's formula, which is obvious when you see it, makes it even clearer.
 
  • Like
Likes vanhees71

What is trigonometric substitution?

Trigonometric substitution is a method used in calculus to simplify and solve integrals involving algebraic expressions and trigonometric functions.

When is trigonometric substitution used?

Trigonometric substitution is used when the integral involves expressions with a square root, or when there is a sum or difference of squares.

How does trigonometric substitution work?

Trigonometric substitution involves replacing the variable in the integral with a trigonometric function, such as sine, cosine, or tangent, and then using trigonometric identities to simplify the integral.

What are the common trigonometric substitutions?

The most commonly used trigonometric substitutions are:
- x = a sinθ
- x = a cosθ
- x = a tanθ
where a is a constant and θ is a variable.

What are the benefits of using trigonometric substitution?

Trigonometric substitution can help simplify complex integrals and make them easier to solve. It also allows for the use of trigonometric identities to simplify the integral and find the solution more efficiently.

Similar threads

Replies
3
Views
938
  • Calculus
Replies
6
Views
1K
Replies
14
Views
1K
Replies
3
Views
218
Replies
6
Views
1K
Replies
21
Views
696
Replies
4
Views
192
  • Calculus
Replies
6
Views
1K
Back
Top