HHow do I solve this integral using substitution?

In summary, the conversation is about solving the integral \int\frac{dx}{\sqrt{x(1-x)}} using substitution, specifically u = \sqrt{x}. The process involves finding the derivative of arcsin(2x-1), which is \frac{2}{\sqrt{4x^2 - 4x + 2}}.
  • #1
bross7
91
0
I'm stuck on how to advance further on this problem and if anyone can point my in the right direction I would be greatly appreciative.

[tex]\int\frac{dx}{\sqrt{x(1-x)}}[/tex]

The integral has to be solved using substitution, but we are required to use
[tex]u=\sqrt{x}[/tex]

From this:
[tex]du=\frac{dx}{2\sqrt{x}}[/tex]

But I am stuck on how to convert the remaining portion of the function in terms of du.
[tex]\int\frac{dx}{u\sqrt{1-x}}[/tex]
 
Physics news on Phys.org
  • #2
I gave it a try and couldn't get anywhere with it. Maple says the answer is arcsin(2x-1).

Is that exactly how the question was given?
 
  • #3
[tex]u = \sqrt{x}[/tex]

so

[tex]u^2 = x[/tex]

and

[tex]2du = \frac{dx}{\sqrt{x}}[/tex]

First use the third equation, then use the second equation to get rid of any other instances of x that're left.

And Shawn is not correct in his solution.

--J
 
Last edited:
  • #4
Shaun's solution looks good to me, what do you propose the actual answer is Justin?
 
  • #5
Complete the square within the square root in the denominator and the apply the result

[tex]\int\frac{dx}{\sqrt{a^2-x^2}} = arcsin\frac{x}{a} [/tex]

spacetime
www.geocities.com/physics_all
 
  • #6
[tex]\int\frac{dx}{\sqrt{x(1-x)}} = 2 \arcsin{\left(\sqrt{x}\right)}[/tex]

Differentiate it and you'll get the integrand.

The derivative of arcsin(2x-1) is [tex]\frac{2}{\sqrt{4x^2 - 4x + 2}}[/tex].

--J
 

FAQ: HHow do I solve this integral using substitution?

What is integration by substitution?

Integration by substitution is a technique used in calculus to solve integrals. It involves replacing the variable in the integrand with a new variable, known as the substitution variable, and then evaluating the integral using the new variable.

When should integration by substitution be used?

Integration by substitution should be used when the integrand contains a product of two functions, one of which is the derivative of the other. This technique is also useful when the integrand contains a nested function.

How do you perform integration by substitution?

To perform integration by substitution, follow these steps:1. Identify a suitable substitution variable, usually denoted by u.2. Rewrite the integrand in terms of the substitution variable.3. Differentiate the substitution variable and multiply it by the differential of the variable inside the integral.4. Substitute the new expression for the integrand in terms of the substitution variable into the integral.5. Integrate the new expression with respect to the substitution variable.6. Substitute the original variable back into the final expression.

What are the benefits of using integration by substitution?

Integration by substitution can simplify complicated integrals and make them easier to solve. It also allows for the use of basic integration rules, such as the power rule, to solve integrals that would be otherwise difficult to evaluate.

Are there any limitations to integration by substitution?

Integration by substitution can only be used for certain types of integrals, specifically those that can be rewritten in terms of the substitution variable. It may also require multiple substitutions and can be time-consuming for more complex integrals.

Similar threads

Replies
3
Views
444
Replies
19
Views
997
Replies
6
Views
551
Replies
6
Views
2K
Replies
25
Views
2K
Replies
12
Views
1K
Replies
8
Views
584
Replies
6
Views
635
Back
Top