Help with IntegralS Very Important Quick

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    Important Integrals
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Discussion Overview

The discussion revolves around a student seeking help with ten integration problems as part of their homework. The student has missed classes due to illness and is looking for guidance on how to approach these integrals, which cover various techniques in integration.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory
  • Debate/contested

Main Points Raised

  • The student presents ten integrals and expresses urgency due to their missed lessons.
  • Some participants emphasize the importance of showing work before receiving help.
  • There are suggestions for specific techniques, such as integration by substitution for certain integrals.
  • Participants provide hints and encouragement, reminding the student to stay calm and take their time.
  • Discussions arise about the correctness of the student's attempts and the methods used, with some participants pointing out mistakes and offering corrections.
  • Different participants suggest various approaches for specific integrals, including trigonometric substitutions and integration by parts.
  • There is a mix of supportive comments and technical guidance, with some participants expressing frustration over the student's lack of understanding of certain concepts.

Areas of Agreement / Disagreement

Participants generally agree on the need for the student to show their work and engage with the problems. However, there are multiple competing views on the methods to solve the integrals, and the discussion remains unresolved regarding the best approaches for each integral.

Contextual Notes

Some participants note that the student has missed significant lessons, which may affect their understanding of integration techniques. There are also indications of varying levels of familiarity with integration among the participants, leading to different suggestions and corrections.

Who May Find This Useful

This discussion may be useful for students struggling with integration techniques, particularly those who have missed classes or are seeking peer support in understanding complex mathematical concepts.

Student from UA
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Hi! I`m poor student from Ukraine! I`m writing u becouse u are my last chans!
So I need to do homewofk for monday-tuesday. I have problem becouse i was ill and siting at home during lessons about integration! Sorry for My bad English, but i hope u understad.
So i have 10 integrals! At first i search in intarnet some calcus. I found www.cal101.com But it give me only result. Function step by step only for money! But i haven`t! what i must do? =( Thanks God and Google I find U.
I hope u help me becouse it`s very very important to me.
So 10 integrals(all indefine):
1) (2*(x^(1/3))+3*x*(x^(1/2)))/(x^(1/4))
2) ((x^3+1)^(1/4))*(x^2)
3) (1-2*sinx)/(cosx^2)
4) 2^(x/2)
5) dx/((4-x^2)^(1/2))
6) tg(x^(1/2))*(dx/x^(1/2))
7) dx/((4+x^2)^(1/2))
8) (sin(x/2))^2
9) lnx/(x^(1/2))
10) (1-3x)cos2x
They are in Attach Files!
 

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Hi and welcome to the forums. Firstly please note that the forum rules state that you must show some work before we can help. What do you know about integration? Have you had a go at any of the questions?

Also, in future, please note that we have homework forums specifically for these type of question.
 
cristo said:
Hi and welcome to the forums. Firstly please note that the forum rules state that you must show some work before we can help. What do you know about integration? Have you had a go at any of the questions?

Also, in future, please note that we have homework forums specifically for these type of question.

I agree with u, - I must do homework not u. I`ve just start so in atach u can see 1 an 2 . But i think it isn`t right/
 

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1. [tex]x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}[/tex]

2. Your technique is incorrect. Have you come across integration by substitution? Try the substitution y=x^3+1
 
U know if i weren`t ill all 2 week, i do home work well. I`m have good marks 87 of 100. I have never ask people to help me in algebra.. butt know. very X situation. I haven`t free time to learn intaegrals, becouse now i do years work in discrete Mathematic. =( Next week i must to complete. Thanks God with discrete mathematica i`m good.
 
cristo said:
1. [tex]x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}[/tex]

2. Your technique is incorrect. Have you come across integration by substitution? Try the substitution y=x^3+1

O fu... Really. It`s child mistake... Oh... Sorry. Hm . I`m trying again.
 
Ok. Don't worry, and don't let youself get stressed; it's only homework, not a life and death situation! Try and calm down a little, and take your time. Then you're less likely to make mistakes.
 
About 1. I`m stuck in a middle. I have result (72x^(13/12))+(52x^(9/4)))/39. Is it right?
About 2. I don`t under stand explain me please.
 
You know i do right the 1st =) I compare it with cal101 result. Cal101 make other version but in result we have right rsult =) He he I love integrals!
 
  • #10
But.. Remain else 9 . What to do?
 
  • #11
ANd u now... U have 3 pm. But i`m in 0.13 AM =( Really want go to sleep but How? =( when i think, what will hapen with me if i will not complete the task...
 
  • #12
About the 2. In calc101.com result (atach file) diferent only with x^3 (see my version) . So where i do mistake , and in the and have misterios x^3
 

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  • #13
How did you get in this bad situation?
 
  • #14
I much powerfully was ill, and has missed nearly 2 weeks lessons. Beside me was a temperature from 35.6 before 39. ANd i couldn`t do anything.
 
  • #15
Right: question 1. [tex]\int\frac{2x^{1/3}+3x^{3/2}}{x^{1/4}}dx=\int 2x^{1/2}+3x^{5/4} dx[/tex] Can you integrate this using the rule for fractional powers?

question 2. [itex]\int(x^3+1)^{1/4}x^2 dx[/itex].
Are you familiar with the method of substitution? If so use the substitution y=x^3+1; this gives dy=3x^2dx. Substituting into the integrand gives [tex]\int y^{1/4}\frac{dy}{3}[/tex]. Can you solve this? The substituting in for x will give you the solution.
 
  • #16
A few more hints for the ones I can see how to do immediately!

5&7 will be some trig subsitutions. For 5 let x=2sinu, for 7 let x=2tanu.
8. looks like it will be do-able using the double angle formula to express it in terms of sinx (or cosx)
10. use integration by parts.

That should give you a bit to work on, I'll have a look at the others in more detail when I've got time [or if anyone else is reading, feel free to help, of course!]
 
  • #17
How much do you know about integration?
 
  • #18
To Crosson:
Not much. =(. But I trying to rescue the situation.
I`ve just get up. Now i start to do the second. And when i finish it, try to make 5&7.
 
  • #19
THX . I `ve done the second! Integration is interesting ... but.. now i just start to do 5... =9
 
  • #20
Hm see the 5. I can`t stand, what i do wrong. help:confused:
 

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  • #21
Hm.. how u get 2sinudu ? if x=2sinu.
 
  • #22
dx=-2cosudu.
 
  • #23
Student from UA said:
Hm.. how u get 2sinudu ? if x=2sinu.

You use the chain rule: dx=d(2sin u)=2 cosu du.

[Sorry, above I said that your notation d(2sin u) was not correct. Of course, it is correct, but not all that useful here!]
 
  • #24
He. the 8 i `ve done for 20-40 sec. =) see in atach
 

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  • #25
The correct integral should be something like;

[tex]x = 2\sin(u) \Rightarrow dx = 2\cos(u) \; du[/tex]

[tex]\int\frac{dx}{\sqrt{4-x^2}} = \int\frac{2\cos(u)}{2\sqrt{1-\sin^2(u)}}du[/tex]

Which should be somewhat easier to integrate.

Edit: Damn beaten to, you get up too early in the morning cristo!
 
Last edited:
  • #26
so how i must write in copy book about the 5 ? how u get 2sinudu can u write?
 
  • #27
Hootenanny said:
The correct integral should be something like;

[tex]x = 2\sin(u) \Rightarrow dx = 2\cos(u) \; du[/tex]

[tex]\int\frac{dx}{\sqrt{4-x^2}} = \int\frac{2\cos(u)}{2\sqrt{1-\sin^2(u)}}du[/tex]

Which should be somewhat easier to integrate.

Edit: Damn beaten to, you get up too early in the morning cristo!

[tex]\int\frac{dx}{\sqrt{4-x^2}} = \int\frac{2\cos(u)}{2\sqrt{1-\sin^2(u)}}du = \int\1du=u+c[/tex] And what to do?
 
  • #28
Hootenanny said:
Edit: Damn beaten to, you get up too early in the morning cristo!

Yea, that's true, but I made a rather stupid mistake. I've deleted my incorrect post, to avoid confusion.

Student: Look at Hootenanny's post for qn 5.
 
  • #29
Student from UA said:
[tex]\int\frac{dx}{\sqrt{4-x^2}} = \int\frac{2\cos(u)}{2\sqrt{1-\sin^2(u)}}du = \int\1du=u+c[/tex] And what to do?
Well, you were given the integrand in terms of x, but now we have an integral in terms of u. So the next step would be to change the integral from a function of u back to a function of x.
 
  • #30
Hootenanny said:
Well, you were given the integrand in terms of x, but now we have an integral in terms of u. So the next step would be to change the integral from a function of u back to a function of x.

U know... I know that. I don`t know how to do that! =(:confused: :cry:
 

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