Integration of √(1-x^2) with x=sinθ: Check Correctness and Simplification

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In summary: It makes it unlikely that you will be able to learn anything. You were given excellent advice to differentiate your answer to see if it was correct. You failed to do so and came back with an inane response. I suggest you apologize to those people who were trying to help you.In summary, the conversation was about determining the correctness of an integration problem involving the expression ##\int \sqrt{1-x^2}dx##. The method of substitution using ##x = \sin \theta## was discussed, and the final result was found to be ##\frac{1}{2}\sin^{-1}x + \frac{1}{2}x\sqrt{1-x^2} + C##
  • #1
askor
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Does my below integration is correct?

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

Let ##x = \sin \theta##, then
##dx = \cos \theta \ d \theta##,
##\cos \theta = \sqrt{1 - x^2}##,
##\theta = \sin^{-1} (x)##

##\int \sqrt{1 - x^2} \ dx##
##= \int \left( \sqrt{1 - \sin^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)##
##= \int \left( \sqrt{\cos^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)##
##= \int \cos \theta \cos \theta \ d\theta##
##= \int \cos^2 \theta \ d\theta##
##= \int \frac{1 + \cos 2\theta}{2} \ d\theta##
##= \frac{1}{2} \int d\theta + \frac{1}{2} \int \cos 2\theta \ d\theta##
##= \frac{1}{2} \theta + \frac{1}{2} \frac{1}{2} \sin 2\theta + C##
##= \frac{1}{2} \theta + \frac{1}{4} \sin 2\theta + C##
##= \frac{1}{2} \theta + \frac{1}{4} 2 \sin \theta \cos \theta + C##
##= \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta + C##
##= \frac{1}{2} \sin^{-1} x + \frac{1}{2} x \sqrt{1 - x^2} + C##

Does this correct?
 
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  • #2
askor said:
Does my below integration is correct?

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

Let ##x = \sin \theta##, then
##dx = \cos \theta \ d \theta##,
##\cos \theta = \sqrt{1 - x^2}##,
##\theta = \sin^{-1} (x)##

##\int \sqrt{1 - x^2} \ dx##
##= \int \left( \sqrt{1 - \sin^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)##
##= \int \left( \sqrt{\cos^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)##
##= \int \cos \theta \cos \theta \ d\theta##
##= \int \cos^2 \theta \ d\theta##
##= \int \frac{1 + \cos 2\theta}{2} \ d\theta##
##= \frac{1}{2} \int d\theta + \frac{1}{2} \int \cos 2\theta \ d\theta##
##= \frac{1}{2} \theta + \frac{1}{2} \frac{1}{2} \sin 2\theta + C##
##= \frac{1}{2} \theta + \frac{1}{4} \sin 2\theta + C##
##= \frac{1}{2} \theta + \frac{1}{4} 2 \sin \theta \cos \theta + C##
##= \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta + C##
##= \frac{1}{2} \sin^{-1} x + \frac{1}{2} x \sqrt{1 - x^2} + C##

Does this correct?
Yes, in case ##|x|\leq 1.##
 
  • #3
fresh_42 said:
Yes, in case ##|x|\leq 1.##

Why ##|x|\leq 1.##?
 
  • #4
askor said:
Why ##|x|\leq 1.##?
Otherwise, it won't be the sine of an angle.

The formula for ##|x|\geq 1## involves ##\cosh^{-1}## at least if the integral table on Wikipedia where I looked it up is correct.
 
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  • #5
If [itex]x > 1[/itex] the integral is [itex]i\int \sqrt{x^2 - 1}\,dx[/itex] and [itex]x = \cosh u[/itex] with [itex]dx = \sinh u\,du[/itex] and [itex]\cosh^2 u - 1 =\sinh^2 u[/itex] is simplest. If [itex]x < -1[/itex] then use [itex]x = -\cosh u[/itex] instead.
 
  • #6
askor said:
Does my below integration is correct?

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

Let ##x = \sin \theta##, then
##dx = \cos \theta \ d \theta##,
##\cos \theta = \sqrt{1 - x^2}##,
##\theta = \sin^{-1} (x)##

##\int \sqrt{1 - x^2} \ dx##
##= \int \left( \sqrt{1 - \sin^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)##
##= \int \left( \sqrt{\cos^2 \theta} \right) \ \left( \cos \theta \ d\theta \right)##
##= \int \cos \theta \cos \theta \ d\theta##
##= \int \cos^2 \theta \ d\theta##
##= \int \frac{1 + \cos 2\theta}{2} \ d\theta##
##= \frac{1}{2} \int d\theta + \frac{1}{2} \int \cos 2\theta \ d\theta##
##= \frac{1}{2} \theta + \frac{1}{2} \frac{1}{2} \sin 2\theta + C##
##= \frac{1}{2} \theta + \frac{1}{4} \sin 2\theta + C##
##= \frac{1}{2} \theta + \frac{1}{4} 2 \sin \theta \cos \theta + C##
##= \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta + C##
##= \frac{1}{2} \sin^{-1} x + \frac{1}{2} x \sqrt{1 - x^2} + C##

Does this correct?
One way to find out is to take the derivative. What does that give you?

-Dan
 
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  • #7
askor said:
Does this correct?

topsquark said:
One way to find out is to take the derivative. What does that give you?
@askor, this is something you should always do in an integration problem. If you differentiate your answer and get back to the original integrand, you know your work is correct (assuming you didn't make any mistakes in differentiating the answer).
 
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  • #8
Can I modify the result such as ##\theta = \cos^{-1} (\sqrt{1 - x^2})##,
so that the final result will be ##\frac{1}{2} \cos^{-1} (\sqrt{1 - x^2}) + \frac{1}{2} x \sqrt{1 - x^2} + C##?
 
  • #9
Hello, anybody home? Can someone please answer my question?
 
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  • #10
askor said:
Hello, anybody home? Can someone please answer my question?

askor said:
Can I modify the result such as ##\theta = \cos^{-1} (\sqrt{1 - x^2})##,
so that the final result will be ##\frac{1}{2} \cos^{-1} (\sqrt{1 - x^2}) + \frac{1}{2} x \sqrt{1 - x^2} + C##?

Mark44 said:
@askor, this is something you should always do in an integration problem. If you differentiate your answer and get back to the original integrand, you know your work is correct (assuming you didn't make any mistakes in differentiating the answer).
https://www.wolframalpha.com/input?i=d/dx+((1/2)*(cos^{-1}+(root(1+-+x^2)+))+++(1/2)+*x+*root(1-x^2))=
 
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  • #11
What does it mean? I don't understand.
 
  • #12
What does what mean?
What don't you understand?
Have you differentiated your suggestion as recommended?
What did you mean in post #8?
Where are your calculations?
Do you know WolframAlpha?
Did you click the link?
What from the link didn't you understand?

Listen, you cannot come here and expect people to do your work or to guess what you might have meant.
 
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  • #13
Observe that for ##x \gt 1## the integrand is complex.
$$
I=\int \sqrt{1-x^2}dx=i\int \sqrt{x^2-1}dx
$$
If you want to go further with this; let ##x=\cosh(u)##, ##dx=\sinh(u)du##.
$$
\sqrt{\cosh^2(u) -1}=\sinh(u)
$$
$$
I=i\int \sinh^2(u)du
$$
$$
\sinh^2(u)=\frac{1}{2}(\cosh(2u)-1)
$$
$$
I=\frac{i}{2}\int (\cosh(2u)-1)du=\frac{i}{4}(\sinh(2u)-2u) + C
$$
$$
=\frac{i}{4}(\sinh(2\cosh^{-1}(x)) -2\cosh^{-1}(x)) + C
$$
$$
=\frac{i}{4}(\sinh(2\ln(x+\sqrt{x^2 -1})) -2\ln(x+\sqrt{x^2 -1})) + C
$$
I advise you not to be so petulant when people are trying to help you.
 
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1. What is the purpose of integrating √(1-x^2) with x=sinθ?

The purpose of integrating √(1-x^2) with x=sinθ is to find the area under the curve of the function √(1-x^2) when x is equal to sinθ. This type of integration is commonly used in calculus and physics to solve problems involving curved shapes and motion.

2. How do you check the correctness of the integration?

To check the correctness of the integration, you can use the fundamental theorem of calculus, which states that the derivative of the integral of a function is equal to the original function. In this case, you can take the derivative of the integrated function and see if it matches with √(1-x^2).

3. What are the steps to simplify the integration of √(1-x^2) with x=sinθ?

The steps to simplify the integration of √(1-x^2) with x=sinθ are as follows:

  1. Use the substitution method to replace x with sinθ.
  2. Apply the Pythagorean identity sin^2θ + cos^2θ = 1 to simplify the expression.
  3. Integrate the simplified expression using the power rule for integration.
  4. Replace sinθ with x in the final result.

4. Can this integration be solved using other methods?

Yes, this integration can also be solved using trigonometric substitution or by converting it into a definite integral and using the substitution method. However, the method of integration with x=sinθ is the most efficient and straightforward approach.

5. What are some real-world applications of integrating √(1-x^2) with x=sinθ?

Integrating √(1-x^2) with x=sinθ has various real-world applications, including calculating the arc length of a circle, finding the area under a curved bridge, and determining the displacement of an object moving in a circular path. It is also used in physics to analyze the motion of objects on a curved path.

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