I don't understand this integral

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In summary, the conversation is about understanding how to integrate a given equation and eliminating variables in the process. The expert suggests using a substitution method and provides a step-by-step process for solving the equation. There is some confusion about the work and the use of substitution, but it is eventually resolved.
  • #1
jkh4
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I don't understand for integral ysin(xy)dx = -cos(xy) for a=1 b=2. I know sin's integral is cos, but I don't understand how to eliminated the y in the left equation so it become the right equation. Please help!
 
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  • #2
What is the derivative to -cos(xy)?

That should help to where the y is going.
 
  • #3
jkh4 said:
I don't understand for integral ysin(xy)dx = -cos(xy) for a=1 b=2. I know sin's integral is cos, but I don't understand how to eliminated the y in the left equation so it become the right equation. Please help!

I am assuming that the a and b are the limits of integration.
And I am assuming that y is a constant here (it's independent of x). Then this is the simplest type of substitution: just define a new variable z=xy. What is dx then? You should then integrate easily (watch out about changing the limits of integration though if you leave your answer in terms of z).
 
  • #4
what about this one?

how do you integrate x(y^2 - x^2)^(1/2)? my TA says the answer is (-1/3)((y^2 - x^2)^(3/2)) but i don't get where is the (-1/3) comes from...
 
  • #5
jkh4 said:
what about this one?

how do you integrate x(y^2 - x^2)^(1/2)? my TA says the answer is (-1/3)((y^2 - x^2)^(3/2)) but i don't get where is the (-1/3) comes from...

Did you not do any substitution rules or anything?

Where is the work for this? Follow the work and it should be clear where it came from.
 
  • #6
this is the process i got so far

(x^2/2)((y^2-x^2)^(3/2))/(3/2)(-1/x^2)

but one thing i don't understand, for the (-1/X^2), is this a proper intergral step?
 
  • #7
jkh4 said:
this is the process i got so far

(x^2/2)((y^2-x^2)^(3/2))/(3/2)(-1/x^2)

but one thing i don't understand, for the (-1/X^2), is this a proper intergral step?

What?

Where does all this come from?
 
  • #8
nevermind , i got it
 
Last edited:
  • #9
One thing that was causing confusion throughout this thread- it was never stated that the integration was to be done with respect to x!
 

1. How do I solve this integral?

Solving integrals can be challenging, but there are several methods that can help. You can use integration by parts, substitution, or partial fractions to solve the integral. It's also important to carefully consider the limits of integration and any known properties of the function.

2. What is the purpose of finding the integral?

The integral is a fundamental concept in calculus that represents the area under a curve. It has many applications in physics, engineering, and other sciences. It can also be used to find important quantities such as volume, work, and displacement.

3. How do I know which method to use for solving the integral?

Choosing the right method for solving an integral depends on the form of the function and any known properties. It's important to understand the different methods and their applications in order to make an informed decision. Practice and experience also play a role in developing this skill.

4. What if I can't solve the integral?

If you are unable to solve an integral, there are several resources available to help. You can consult with a tutor or professor, use online tools and calculators, or refer to textbooks and other reference materials. It's also important to review the fundamentals of calculus and practice solving simpler integrals before attempting more complex ones.

5. Is it possible to solve this integral without using calculus?

In some cases, it may be possible to solve an integral without using calculus. This can be done by using geometric techniques, such as finding the area of a shape, or by using properties of the function to simplify the integral. However, calculus is often the most efficient and accurate method for solving integrals.

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