STRACT: How to solve a tricky integration problem using multiple substitutions?

In summary: Done!In summary, the student attempted to solve an indefinite integral with the help of a trig substitution, but ran into trouble when trying to solve for x. After making a few more substitutions, they were finally able to get the answer.
  • #1
2h2o
53
0

Homework Statement



Evaluate the indefinite integral:

Homework Equations



[tex]\int\frac{6}{x\sqrt{25x^2-1}} dx[/tex]

The Attempt at a Solution



I've tried a number of different substitutions, but this last one gets me the closest, I think.

[tex]\int\frac{6}{x\sqrt{25x^2-1}} dx = \frac{6}{5}\int\frac{1}{x\sqrt{x^2-\frac{1}{25}}} dx = \frac{6}{5}\int\frac{x}{x^2\sqrt{x^2-\frac{1}{25}}} dx[/tex]


let [tex]u=\sqrt{x^2-\frac{1}{25}[/tex]

then

[tex]du=\frac{x dx}{\sqrt{x^2-\frac{1}{25}}}}}[/tex]

This is where I want to substitute in, but it doesn't go. [tex]\frac{x}{\sqrt{x^2-\frac{1}{25}}}}}}dx[/tex] fits in, except for the x^2 in the denominator which doesn't work with my substitution. I don't see where to go from here, or what I should have done differently.

Thanks!
 
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  • #2
Try a trig substitution x=cosecθ
 
  • #3
2h2o said:

Homework Statement



Evaluate the indefinite integral:

Homework Equations



[tex]\int\frac{6}{x\sqrt{25x^2-1}} dx[/tex]

The Attempt at a Solution



I've tried a number of different substitutions, but this last one gets me the closest, I think.

[tex]\int\frac{6}{x\sqrt{25x^2-1}} dx = \frac{6}{5}\int\frac{1}{x\sqrt{x^2-\frac{1}{25}}} dx = \frac{6}{5}\int\frac{x}{x^2\sqrt{x^2-\frac{1}{25}}} dx[/tex]


let [tex]u=\sqrt{x^2-\frac{1}{25}[/tex]

then

[tex]du=\frac{x dx}{\sqrt{x^2-\frac{1}{25}}}}}[/tex]

This is where I want to substitute in, but it doesn't go. [tex]\frac{x}{\sqrt{x^2-\frac{1}{25}}}}}}dx[/tex] fits in, except for the x^2 in the denominator which doesn't work with my substitution. I don't see where to go from here, or what I should have done differently.

Thanks!

You're almost done: after substituting x^2=u^2+1/25 in the last integral you obtained, just bring 5 back into the integrand and then you need another substitution, basically, think of u=\sqrt(5)*cosh(x). Then you'll be left with a simple integral which is neccessarily required to be given a form like e^t/(1+e^2t) dt. From this point, you need one more substitution and then you are done by going back to x as you begin from this last one, pass t and u to definitely hit x again.

AB
 

1. What is integration and why is it important in Calc II?

Integration is a mathematical process used to find the area under a curve or the accumulation of a quantity over a given interval. It is important in Calc II because it allows us to solve a wide range of real-world problems, such as finding volumes, areas, and work done.

2. What are the main techniques used in integration problems?

The main techniques used in integration problems are substitution, integration by parts, partial fractions, and trigonometric substitution. These techniques help simplify complex integrals and allow us to solve them more easily.

3. How do I know which integration technique to use?

Choosing the right integration technique depends on the form of the integral you are trying to solve. Some techniques work better for certain types of integrals, so it is important to familiarize yourself with each technique and practice using them to determine which one is best for a particular problem.

4. What is the difference between indefinite and definite integration?

Indefinite integration is the process of finding a general solution to an integral, while definite integration involves finding the value of an integral within a specified interval. In other words, indefinite integration results in a function, whereas definite integration results in a numerical value.

5. How can I check my answers when solving integration problems?

You can check your answers by differentiating the solution and comparing it to the original integrand. If the resulting expression is the same as the original, then your answer is correct. You can also use online calculators or graphing tools to visualize the area under the curve and confirm your answer.

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