Integral Calculation - Get Help Solving Problems

  • Thread starter cataldo
  • Start date
  • Tags
    Integral
In summary, the conversation is about a problem with a specific integral that the speaker is struggling to solve. They have tried using substitution and online calculators, but the integral is still too complicated. Another person suggests a different substitution and the speaker is unsure if their result is correct. They also mention that the software Mathematica is unable to solve the definite integral and may require numerical integration.
  • #1
cataldo
25
0
I have some problem with following integral. In attachment a pdf with integral..
I have done a integration by sobstitution but i have not a result ...
Tank for your in interest
best regards
 

Attachments

  • integral.pdf
    41.4 KB · Views: 250
Physics news on Phys.org
  • #2
why couldn't you type that out? :frown:

try substituting so that sin becomes cos of something else.
 
  • #3
Ok...
I have done this question because the calculate on site "wolfram mathematica online integrator" only indefinite integral is very complicated and i don't understand as i can obtain this...
Cataldo...
 

Attachments

  • Wolfram Mathematica Online Integrator.pdf
    37.1 KB · Views: 486
  • #4
By a little manipulation you can transform it into an integral from 0 to 1 of the following:

2[(1+x)1/2 + [(1-x)1/2]/(1-x2). Essentially this was done by collapsing the integral into 0 to pi/2 and then letting x=sin.

See if you can go further.
 
  • #5
Ok...Tanks mathman
Is exactly x=sin and integral from 0 to 1.. but i obtain [(1+x)^(1/2)]/(1-x^2)... Does it correct?
 
  • #6
tiny-tim said:
why couldn't you type that out? :frown:

try substituting so that sin becomes cos of something else.

how about sinx = cos(π/2 - x)? :wink:
 
  • #7
cataldo said:
Ok...Tanks mathman
Is exactly x=sin and integral from 0 to 1.. but i obtain [(1+x)^(1/2)]/(1-x^2)... Does it correct?

The sin is positive from 0 to pi and negative from pi to 2pi. I first collapsed the second interval into the first, so I had two terms: 1+sin and 1-sin. Next collapsing to 0 to pi/2 with a factor of 2. Finally let x=sin.
 
  • #8
Hi nathaman..Tanks for your replay
I understand your method but the software mathematica can not assess the definte integral because the fuction between 0 to 1 not converge...exist the indefinte integral indeed mathematica assess it... I think that is necessary assess it by numerical integration...
 

FAQ: Integral Calculation - Get Help Solving Problems

What is integral calculus?

Integral calculus is a branch of mathematics that deals with the calculation of areas and volumes, as well as the determination of the rate of change of a function over a specific interval.

What is the purpose of integral calculus?

The main purpose of integral calculus is to find the total accumulated quantity or value of a changing quantity over a specific interval, such as finding the total distance traveled by an object given its velocity function.

What are the different types of integrals?

There are two types of integrals in integral calculus: definite integrals and indefinite integrals. A definite integral is used to find the exact area under a curve between two points, while an indefinite integral is used to find a general function whose derivative is equal to the given function.

How do I solve an integral?

To solve an integral, you can use several methods such as integration by substitution, integration by parts, or using specific formulas for trigonometric, exponential, or logarithmic functions. It is important to have a good understanding of the fundamental principles of integral calculus before attempting to solve integrals.

Why is it important to learn how to solve integrals?

Integrals are essential in many fields of science and engineering, including physics, chemistry, and engineering. They are used to solve real-world problems such as finding the area under a curve, calculating volumes, and determining the amount of work done by a force. Additionally, integral calculus is the basis for more advanced mathematical concepts such as differential equations.

Similar threads

Back
Top