Trouble with Integrating (1/x^2)*sec(1/x)*tan(1/x)dx Using u-Substitution

  • Thread starter fiziksfun
  • Start date
  • Tags
    Integral
In summary, the conversation discusses the use of u-substitution in solving integrals and the confusion surrounding the substitution of u^2 in the original integral. The expert explains that u should be chosen so that du is equal to a function that is already inside the integral, simplifying the integral in terms of u and du without any x's.
  • #1
fiziksfun
78
0
So this is my first year of calculus and we're doing integrals using u-substitution

I'm having trouble with this integral:

integral of (1/x^2)*sec(1/x)*tan(1/x)dx

i let u=(1/x)
so du=(1/x^2)dx
so du(x^2)=dx

If I do this, the integral works out perfectly since the x^2 cancel out

so it equals sec(u) + C

HOWEVER, my question is
if i let u = (1/x)

why don't i replace the (1/x^2) (in the beginning of the integral) with u^2

CAN SOMEONE HELP ME??
 
Physics news on Phys.org
  • #2
First of all, the derivative of [tex]\frac{1}{x}[/tex] is not [tex]\frac{1}{x^2}[/tex]


Second of all, if [tex]u^2 = \frac{1}{x^2}[/tex]


then [tex]2udu = \frac{-2}{x^3}dx[/tex]


And you haven't made anything simpler for yourself.
 
  • #3
the derivative of 1/x is -1/x^2 correct? sorrrybut if i were to do a u substitution like

integral of tan^4(x)*sec^2(x)

and i let u=tan x
then du*sec^2x=dxand that leads to the integral of u^4, correct? I'm confused
 
  • #4
I don't follow. The derivative of u with respect to x is du / dx and the derivative of tan x is (secx)^2. so du / dx = (secx)^2.
 
  • #5
right. so dx = du / (secx)^2 so
if i substitute that back into the original integral the (secx)^2 cancel out.
is the clearer?
 
  • #6
But you can't let u = tan(x) because there is no tan(x) in the integral. There's a tan(1/x).
 
  • #7
okk new problem, just to show what I am trying to say
lets take this integral
http://archives.math.utk.edu/visual.calculus/4/substitutions.1/int1-2.gif

the solution is
http://archives.math.utk.edu/visual.calculus/4/substitutions.1/sint1-2.gif

we let u=cosx

when we substitute it back into the original integral we have u^3

that's what confuses me. why can i do u^3 there and not
u^2 in the my original integral

does that make sense??
 
  • #8
The difference is that you are saying "let u^2 = some function, whereas in this question, we let u = some function, and when you substitute u back into the integral, it just happens to become u^3.

What you want to do is let u = some function so that du = (some function that's already inside the integral)dx. This simplifies things in terms of u and du with no X's.
 

1. What is a "somewhat easy integral"?

A "somewhat easy integral" is a type of mathematical integration problem that falls in between being extremely easy and extremely difficult. It typically involves basic algebraic manipulation and may require the use of integration techniques such as substitution or integration by parts.

2. How do I know if an integral is considered "somewhat easy"?

Determining the difficulty level of an integral is subjective and can vary depending on an individual's mathematical proficiency. However, some indicators of a "somewhat easy integral" may include a relatively simple integrand, a limited number of terms, and the ability to apply common integration techniques.

3. Can you give an example of a "somewhat easy integral"?

One example of a "somewhat easy integral" is ∫(x^2 + 2x + 1)dx. This integral can be solved by using the power rule and basic algebraic manipulation, making it relatively straightforward compared to more complex integrals.

4. What are some tips for solving "somewhat easy integrals"?

First, identify the integration techniques that can be applied to the integral. These may include the power rule, substitution, or integration by parts. Then, carefully manipulate the integrand to simplify the problem. It can also be helpful to graph the integrand to gain a better understanding of the integral's behavior.

5. Are "somewhat easy integrals" important in real-world applications?

Yes, "somewhat easy integrals" are commonly used in various fields of science and engineering to solve problems involving rates of change, areas and volumes, and other physical phenomena. They serve as building blocks for more complex integrals and are essential in understanding the fundamental principles of calculus.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
341
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
491
  • Calculus and Beyond Homework Help
Replies
15
Views
783
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
759
Back
Top