How to Solve Integrals by Substitution: Tips and Tricks

  • Thread starter powp
  • Start date
  • Tags
    Integral
In summary, the problem is that the original integral doesn't seem to be working, and it seems like subsitution might be the solution. However, it's not clear how to do the substitution. Someone else suggested working with u^2 instead of 1-x, and that seems to have worked. x= 1-u^2, and the integral becomes -2u * du= dx.
  • #1
powp
91
0
Hello All

I am having problem with solving integrals by subsitution.

I have the following problem. Can anybody help?

(large S) X^2 /(SQRT(1-X)) DX

My first and only thought was to subsitute 1 - x with U but that gave me an answer not ever close.

Does anybody have any tips?

Thanks

P
 
Physics news on Phys.org
  • #2
powp said:
Hello All
I am having problem with solving integrals by subsitution.
I have the following problem. Can anybody help?
(large S) X^2 /(SQRT(1-X)) DX
My first and only thought was to subsitute 1 - x with U but that gave me an answer not ever close.
Does anybody have any tips?
Thanks
P

Powp, here it is in LaTex:

[tex]\int \frac{x^2}{\sqrt{1-x}}dx[/tex]

So let:

[tex]u=\sqrt{1-x}[/tex]

and work it though completely. That is, then what is [itex]u^2[/itex]?

What then is x in terms of u?

What then is dx in terms of u and du?

What is [itex]x^2[/itex] in terms of u?
 
  • #3
Thanks for the reply

u^2 = 1 - x
2u * du = -1 * dx
-2u * du = dx

x = 1 - u^2

but where do I go from here.
 
  • #4
Put them into the integral and DO it!
Your original integral was
[tex]\int \frac{x^2}{\sqrt{1-x}}dx[/itex]
If you let [itex]u= \sqrt{1-x}= (1-x)^\frac{1}{2}[/itex] then, yes, [itex]u^2= 1-x[/itex] so 2u du= -dx and x= 1- u2 so x2= (1- u2)2= 1- 2u2+ u4. Your integral becomes
[tex]\int \frac{1- 2u^2+ u^4}{u}(-2u du)= -2\int \left(1- 2u^2+ u^4\right)du[/itex].
Actually, your first thought: substituting for 1- x works perfectly well.
If u= 1-x then du= -dx. Also, x= 1- u so x2= (1- u)2= 1- 2u+ u2. The integral becomes
[tex]-\int \frac{1- 2u+ u^2}{u^\frac{1}{2}}du= -\int \left(u^{-\frac{1}{2}}- 2u^\frac{1}{2}+ u^\frac{3}{2}\right) du[/tex]
 
Last edited by a moderator:
  • #5
powp said:
Thanks for the reply
u^2 = 1 - x
2u * du = -1 * dx
-2u * du = dx
x = 1 - u^2
but where do I go from here.

We're makin' progress. Now you need to substitute all these expressions into the original integral:

[tex]\int\frac{x^2}{\sqrt{1-x}}dx[/tex]

So, substitute each one. I'll do one of them:

[tex]dx=-2udu[/tex]

There you go. Substitute -2udu for dx in the integral. Well . . . do the same for [itex]x^2[/itex] and [itex]\sqrt{1-x}[/itex] . . . what does the integral then look like in terms of u? It's a lot easier now right? So solve it for u then don't forget to substitute back [itex]\sqrt{1-x}[/itex] in place of u in the final answer.

Edit: Didn't see Hall's reply but you know what to do now.
 
Last edited:

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to find the total value or quantity of a function over a given interval.

2. Why is evaluating integrals important?

Evaluating integrals is important because it allows us to solve real-world problems and make predictions based on mathematical models. It is also used extensively in fields such as physics, engineering, and economics.

3. How do you evaluate an integral?

To evaluate an integral, you can use various techniques such as the fundamental theorem of calculus, integration by substitution, integration by parts, or numerical methods such as the trapezoidal rule or Simpson's rule.

4. What are the different types of integrals?

There are two main types of integrals: indefinite integrals and definite integrals. Indefinite integrals do not have any limits of integration and represent the antiderivative of a function. Definite integrals have limits of integration and represent the area under a curve between those limits.

5. What is the relationship between derivatives and integrals?

The fundamental theorem of calculus states that the derivative of an integral of a function is equal to the original function. In other words, the derivative and the integral are inverse operations of each other.

Similar threads

  • Calculus and Beyond Homework Help
Replies
24
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
795
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
641
  • Calculus and Beyond Homework Help
Replies
3
Views
267
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
Replies
9
Views
661
  • Calculus and Beyond Homework Help
Replies
5
Views
741
Back
Top