How Do You Solve the Integral of (sin(5x))^(-2) dx?

  • Thread starter Mag|cK
  • Start date
  • Tags
    Integral
In summary, the conversation is about solving the integral of (sin5x)^(-2) dx using substitution methods. The first substitution suggested is u=sin(5x), but this only leads to another difficulty with the cos5x term. The next suggestion is to try the substitution v=1/u. Finally, it is mentioned that using -d[cot(x)]=(sinx)^(-2) dx can easily solve the integral.
  • #1
Mag|cK
36
0
Hello, i have a question. This is integral question, but i don't have any program to write the equation with:

integral (sin5x)^(-2) dx= ?

Thank you..
 
Physics news on Phys.org
  • #2
This is one of those you should really know, since it can be rewritten cosec2(5x).

However, if you don't know the antiderivative, then it can be worked out by a few substitutions, the first being u=sin(5x)
 
  • #3
cristo said:
This is one of those you should really know, since it can be rewritten cosec2(5x).

However, if you don't know the antiderivative, then it can be worked out by a few substitutions, the first being u=sin(5x)

yes i have done using substitution method but can't solve the problem. Can you step by step show me how please? thx
 
  • #4
It's your homework, not mine! Show how you did the first substitution, and we'll go from there.
 
  • #5
cristo said:
It's your homework, not mine! Show how you did the first substitution, and we'll go from there.

ok sorry :smile:
if i substitute with sin5x,

integral (sin5x)^(-2) d(sin5x)/5cos5x

this is how far i can get if i substitute with sin5x as i can't eliminate the cos5x.
I already tried substituting with cosec5x and sec5x, but still no solution.
 
  • #6
With -d[cot(x)]=(sinx)^(-2) dx ,you can solve it easily.Have a try!
 
  • #7
Mag|cK said:
ok sorry :smile:
if i substitute with sin5x,

integral (sin5x)^(-2) d(sin5x)/5cos5x

this is how far i can get if i substitute with sin5x as i can't eliminate the cos5x.
If you've substituted, why do you still have functions of x in the integral? If you let u=sin(5x), then du=5cos(5x)dx=5[sqrt(1-u2)]. This transforms your integral to [tex]5 \int \frac{du}{u^2\sqrt{1-u^2}}[/tex].

There you go, I've done one step for you. Now, try the substitution v=1/u.
I already tried substituting with cosec5x and sec5x, but still no solution.
If you look back, I said you'll need a few substitutions!


jakcn001 said:
With -d[cot(x)]=(sinx)^(-2) dx ,you can solve it easily.Have a try!

I did allude to that in my first post.
 
  • #8
ok thanks guys i understand now :smile:
 

FAQ: How Do You Solve the Integral of (sin(5x))^(-2) dx?

What is an integral?

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

How do I solve an integral?

The process of solving an integral involves finding the antiderivative of a function and then evaluating it at the limits of integration. This can be done using various techniques such as substitution, integration by parts, and trigonometric identities.

What is the purpose of using integrals?

Integrals are used in a wide range of fields such as physics, engineering, economics, and statistics. They help us to find the total value or quantity of a function and are essential in solving real-world problems involving rates of change and accumulation.

Can I use a calculator to solve integrals?

Yes, there are many online calculators and software programs that can solve integrals for you. However, it is important to have a good understanding of the concepts behind integration and how to manually solve integrals to ensure accuracy and avoid errors.

Are there any tips for solving integrals?

One helpful tip for solving integrals is to always check your answer by taking the derivative of the antiderivative you found. This will help to catch any mistakes and ensure that your solution is correct. It is also important to practice and familiarize yourself with different integration techniques to become more efficient at solving integrals.

Similar threads

Back
Top