##\int \frac{1}{\sqrt{1-x^2}} dx##

  • Thread starter Bashyboy
  • Start date
  • Tags
    Dx
In summary, The student tried two different substitution methods to solve the integral, but neither worked. They eventually figured out that the substitution ##\cos \theta=x## was incorrect and that the substitution ##\sin \theta=x## worked.
  • #1
Bashyboy
1,421
5

Homework Statement


I am doing a little review and having a some trouble deriving the integral ##\int \frac{1}{\sqrt{1-x^2}} dx##

Homework Equations

The Attempt at a Solution


Initially I was trying to solve this integral using the substitution ##\cos \theta = x##. I drew my triangle so that the side adjacent to ##\theta## was ##x##, the hypotenuse was ##1##, and from this found that the opposite side was ##\sqrt{1-x^2}##; after this I computed my differentials, performed various substitutions, and concluded that ##\int \frac{1}{\sqrt{1-x^2}} dx = - \arccos x + c##. However, this wrong. So, I tried the substitution ##\sin \theta = x## and I derived the correct formula. Why didn't my first substitution work?
 
Physics news on Phys.org
  • #3
Bashyboy said:

Homework Statement


I am doing a little review and having a some trouble deriving the integral ##\int \frac{1}{\sqrt{1-x^2}} dx##

Homework Equations

The Attempt at a Solution


Initially I was trying to solve this integral using the substitution ##\cos \theta = x##. I drew my triangle so that the side adjacent to ##\theta## was ##x##, the hypotenuse was ##1##, and from this found that the opposite side was ##\sqrt{1-x^2}##; after this I computed my differentials, performed various substitutions, and concluded that ##\int \frac{1}{\sqrt{1-x^2}} dx = - \arccos x + c##. However, this wrong. So, I tried the substitution ##\sin \theta = x## and I derived the correct formula. Why didn't my first substitution work?

Your first expression ##-\arccos(x) + C## is correct; differentiate it to check. That is something that you should always do.

Think about why there are two different-looking answers!
 
  • #4
If you can't figure out what Ray's getting at, try this: set C=0 for simplicity and plot the two results.
 
  • #5
Call the integral ##\theta\equiv \int \frac{1}{\sqrt{1-x^2}} dx##.
Then we have
##\theta=-\arccos(x) + C_1## and ##\theta=\arcsin(x) + C_2##,

Something that is sometimes forgotten is that the constants of integration (##C_1## and ##C_2## ) are not generally equal.
That is, if we choose a value for ##C_1##, what is the corresponding choice for ##C_2##?
Let ##C_2=C_1+Q##, where ##Q## is an unknown.

In this case, it might help interpreting the results as follows:
##\cos(C_1-\theta)=x ## and ##\sin(\theta-C_2)=x##.
What is ##Q##?
 

Related to ##\int \frac{1}{\sqrt{1-x^2}} dx##

1. What is the purpose of the integral ##\int \frac{1}{\sqrt{1-x^2}} dx##?

The purpose of this integral is to calculate the arc length of a circle or a portion of a circle. It can also be used to find the area under the curve of a semicircle or quarter circle.

2. How is the integral ##\int \frac{1}{\sqrt{1-x^2}} dx## solved?

This integral can be solved using a trigonometric substitution, specifically ##x = \sin{\theta}##. By substituting this value and using trigonometric identities, the integral can be simplified and solved.

3. What is the domain of the integral ##\int \frac{1}{\sqrt{1-x^2}} dx##?

The domain of this integral is ##-1 \leq x \leq 1##. This is because the denominator cannot equal 0, and the function is undefined for values of x outside of this range.

4. Can the integral ##\int \frac{1}{\sqrt{1-x^2}} dx## be solved using other methods?

Yes, there are other methods that can be used to solve this integral, such as integration by parts or using a power series expansion. However, the trigonometric substitution method is the most commonly used and efficient method for solving this particular integral.

5. What is the geometric interpretation of the integral ##\int \frac{1}{\sqrt{1-x^2}} dx##?

The geometric interpretation of this integral is the calculation of the length of an arc or the area under a curve of a circle with a radius of 1. It can also be thought of as calculating the distance traveled along a circular path with a radius of 1.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
902
  • Calculus and Beyond Homework Help
Replies
9
Views
851
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
448
  • Calculus and Beyond Homework Help
Replies
3
Views
498
  • Calculus and Beyond Homework Help
2
Replies
44
Views
4K
  • Calculus and Beyond Homework Help
Replies
2
Views
971
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
925
Back
Top