Help with IntegralS Very Important Quick

In summary, I'm poor student from Ukraine. I'm writing you because you are my last hope. I need to do homework for Monday-Tuesday. I have a problem because I was ill and I couldn't do my homework during the lessons about integration. I found www.cal101.com But it only gave me results for money functions. I found www.calc101.com But it said that I needed to do a function step-by-step and that I couldn't use calculus. I found www.uwmath.com But it said that I needed to know the integration by substitution method. But I've never been able to do that.
  • #71
For 10 use integration by parts... Take u=(1-3x) and dv=cos(2x)dx.

Damn, I'm beaten to it this time... by a contributor :approve: . That bandwagons getting pretty full now; I may have to think of jumping on!
 
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  • #72

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  • #73
i mean.. show integration by part becouse not has quite understood
 
  • #74
And what about the 7( " ln (sec u + tan u) + C" if x=2tgu.)?
 
  • #75
Awww, here's 0.24 A.M. =.=" 20h25 is still early, you should stay up a little bit more. :)

7. When seeing some integral in this form:
[tex]\int \frac{dx}{\sqrt{x ^ 2 + \alpha ^ 2}}[/tex], one should think right away about making the trig-substitution [tex]x = \alpha \tan t , \quad t \in \left] -\frac{\pi}{2} ; \ \frac{\pi}{2} \right[[/tex]
It goes like this:
[tex]x = \alpha \tan t \Rightarrow dx = (\alpha \tan t)'_t dt = \alpha \frac{1}{\cos ^ 2 t} dt[/tex]
Your integral will become:
[tex]... = \int \frac{\frac{dt}{\cos ^ 2 t}}{\sqrt{\alpha ^ 2 + \alpha ^ 2 \tan ^ 2 t}} = \int \frac{dt}{ \cos ^ 2 t \sqrt{\alpha ^ 2 \left(1 + \tan ^ 2 t \right)}}[/tex]
[tex]= \int \frac{dt}{\cos ^ 2 t \sqrt{\frac{1}{\cos ^ 2 t}}}[/tex], since [tex]t \in \left] -\frac{\pi}{2} ; \ \frac{\pi}{2} \right[[/tex], so cos t > 0, we have:
[tex]... = \int \frac{dt}{\cos ^ 2 t \sqrt{\frac{1}{\cos ^ 2 t}}} = \int \frac{dt}{\cos ^ 2 x \frac{1}{\cos t}} = \int \frac{dt}{\cos t}[/tex]
Now the power of cosine function is odd (in this case, 1), we'll make the substitution u = sin x, otherwise, when the power of sine function is odd, we make the substitution u = cos x.
[tex]... = \int \frac{\cos t dt}{\cos ^ 2 t} = \int \frac{\cos t dt}{1 - \sin ^ 2 t}[/tex]
Now let u = sin t, du = cos t dt, so the integral becomes:
[tex]... = \int \frac{du}{1 - u ^ 2} = ...[/tex], pretty straightforward from here. Can you go from here? :)

--------------

Number 10, Integration by Parts, you should remember LIATE. :) See post #67. What should u and dv be?

--------------

Number 9, you should do it using [tex]u = \ln (x) \quad dv = x ^ {-\frac{1}{2}} dx[/tex]. It should be easier.

Btw, you've differentiated incorrectly. You still have ln(x) in your du, so it's impossible to do it by letting [tex]u = \ln x x ^ {-\frac{1}{2}} \quad dv = dx[/tex]:
[tex]u = \ln x x ^ {-\frac{1}{2}} \Rightarrow du = \left( (\ln(x))' x ^ {-\frac{1}{2}} + \ln (x) \left( x ^ {-\frac{1}{2}} \right) ' \right) dx[/tex]
[tex]= \left( \frac{1}{x \sqrt x} - \frac{1}{2} \frac{\fbox{\ln(x)}}{x ^ {\frac{3}{2}}} \right) dx[/tex]
 
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  • #76
VietDao29 said:
Number 10, Integration by Parts, remember LIAT. :) See post #67. What should u and dv be?

That's a neat thing to remember. I've never seen that before! (Sorry, I'll stop cluttering the thread now!)
 
  • #77
cristo said:
That's a neat thing to remember. I've never seen that before! (Sorry, I'll stop cluttering the thread now!)

Whooops, I did mistype it. There should be one more E at the end, which means it should read: LIATE, E for exponential. Editing the post now. o:)
 
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  • #78
about 9.. u right i forgot that (uv)`= u`v+v`u
 
  • #79
Student from UA said:
I`ve started... and stop. :bugeye: :confused: What i must do?
http://math2.org/math/integrals/more/ln.htm i`ve read it... Thare is integration by part... And i can`t compare both integrals =( Can u show me on my integral?

[tex]dv = x ^ {- \frac{1}{2}} dx \Rightarrow v = \int dv = \int x ^ {- \frac{1}{2}} dx = 2 \sqrt{x} + C = 2 \sqrt{x}[/tex] (C can be any constant), so we choose C = 0 to make it look more simple.
You can write [tex]dv = x ^ {- \frac{1}{2}} dx \Rightarrow v = 2 \sqrt{x}[/tex] for short. :)
 
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  • #80
about 7: [tex]... = \int \frac{du}{1 - u ^ 2} = ...[/tex] By what formula i must integrate? becouse i can`t find it..
P.S. Maybe nobody in my group don`t know integrals like me now couse for this day i learned so much..
 
  • #81
in 9th result : 2ln(x)sqrtx - 4sqrtx +c right?

Yes right... i`ve just compare with cal101 =) he he =)
 
  • #82
there is 7 and 10 not comleted... agrrr
 
  • #83
Student from UA said:
about 7: [tex]... = \int \frac{du}{1 - u ^ 2} = ...[/tex] By what formula i must integrate? becouse i can`t find it..
P.S. Maybe nobody in my group don`t know integrals like me now couse for this day i learned so much..

[tex]... = \int \frac{du}{(1 - u) (1 + u)} = \frac{1}{2} \int \frac{(1 - u) + (1 + u)}{(1 - u) (1 + u)} du = \frac{1}{2} \int \left( \frac{1 - u}{(1 - u) (1 + u)} + \frac{1 + u}{(1 - u) (1 + u)} \right) du[/tex]

[tex]= \frac{1}{2} \int \left( \frac{1}{1 + u} + \frac{1}{1 - u} \right) du = ...[/tex] You should be able to go from here. :)There are several ways to go about integrating 1/cos(x) = sec(x) (sec(x) is another way to write 1 / cos(x)), this is one of the two common ways, the other way is:
[tex]\int \sec(x) dx = \int \sec(x) \frac{\sec(x) + \tan (x)}{\sec(x) + \tan (x)} dx = \int \frac{\sec ^ 2 x + \sec (x) \tan (x)}{\sec(x) + \tan (x)} dx[/tex]
Let u = sec(x) + tan(x)
[tex]\Rightarrow du = \left( \frac{1}{\cos x} + \tan(x) \right)'_x dx = \left( \frac{\sin (x)}{\cos ^ 2 (x)} + \frac{1}{\cos ^ 2 x} \right) dx = \left( \sec(x) \tan(x) + \sec ^ 2 (x) \right) dx[/tex]
The integral will become:
[tex]\int \frac{du}{u} = \ln|u| + C = \ln |\sec(x) + \tan(x)| + C[/tex]
 
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  • #84
VietDao29 u are the monster of Integrals! My teacher is stupid if her compare with u.
 
  • #85
VietDao29 said:
[tex]... = \int \frac{du}{(1 - u) (1 + u)} = \frac{1}{2} \int \frac{(1 - u) + (1 + u)}{(1 - u) (1 + u)} du = \frac{1}{2} \int \left( \frac{1 - u}{(1 - u) (1 + u)} + \frac{1 + u}{(1 - u) (1 + u)} \right) du[/tex]

[tex]= \frac{1}{2} \int \left( \frac{1}{1 + u} + \frac{1}{1 - u} \right) du = ...[/tex] You shoule be able to go from here. :)


There are several ways to go about integrating 1/cos(x) = sec(x) (sec(x) is another way to write 1 / cos(x)), this is one of the two common ways, the other way is:
[tex]\int \sec(x) dx = \int \sec(x) \frac{\sec(x) + \tan (x)}{\sec(x) + \tan (x)} dx = \int \frac{\sec ^ 2 x + \sec (x) \tan (x)}{\sec(x) + \tan (x)} dx[/tex]
Let u = sec(x) + tan(x)
[tex]\Rightarrow du = \left( \frac{1}{\cos x} + \tan(x) \right)'_x dx = \left( \frac{\sin (x)}{\cos ^ 2 (x)} + \frac{1}{\cos ^ 2 x} \right) dx = \left( \sec(x) \tan(x) + \sec ^ 2 (x) \right) dx[/tex]
The integral will become:
[tex]\int \frac{du}{u} = \ln|u| + C = \ln |\sec(x) + \tan(x)| + C[/tex]

Agrrr:mad: :grumpy: See post 63! =)[tex]\int \frac{du}{u} = \ln|u| + C = \ln |\sec(x) + \tan(x)| + C[/tex] I have the same but how to solve "[tex]\int \frac{dy}{y} = \ln|y| + C = \ln |\sec(u) + \tan(u)| + C[/tex]" if x=2tgu
 
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  • #86
VietDao29 said:
[tex]= \frac{1}{2} \int \left( \frac{1}{1 + u} + \frac{1}{1 - u} \right) du = ...[/tex] You should be able to go from here. :)

Really? Or i`m full idiot or u are great scientist! =)
I have variant [tex]\int \frac{dy}{y} = \ln|y| + C = \ln |\sec(u) + \tan(u)| + C[/tex] and want to complete it, but i`m interesting in u`r variant of solve too... =)
 
  • #87
Student from UA said:
Agrrr:mad: :grumpy: See post 63! =)[tex]\int \frac{du}{u} = \ln|u| + C = \ln |\sec(x) + \tan(x)| + C[/tex] I have the same but how to solve "[tex]\int \frac{dy}{y} = \ln|y| + C = \ln |\sec(u) + \tan(u)| + C[/tex]" if x=2tgu

Whoops, I didn't read the whole a 6-page thread. Just skim through the main discussion :blushing: :biggrin:.
Ok, if x = 2 tan(u) ~~~> x / 2 = tan(u)
Note that [tex]u \in \left] - \frac{\pi}{2}; \ \frac{\pi}{2} \right[[/tex], so cos u > 0, we have:
[tex]\tan u = \frac{x}{2} \Rightarrow 1 + \tan ^ 2 u = 1 + \frac{x ^ 2}{4} \Rightarrow \sec ^ 2 u = 1 + \frac{x ^ 2}{4} \Rightarrow \sec u = \sqrt{1 + \frac{x ^ 2}{4}}[/tex], so plug it to the expression above, yields:
[tex]... = \ln \left| \sqrt{1 + \frac{x ^ 2}{4}} + \frac{x}{2} \right| + C[/tex]
 
  • #88
cristo said:
Damn, I'm beaten to it this time... by a contributor :approve: . That bandwagons getting pretty full now; I may have to think of jumping on!
:biggrin: Join the club, we're thinking of getting jackets made! The £9.00 is worth it just for the avatar!

Student: You have been told how to change your function back into a function of x already. Substitute in a have a play...
 
  • #89
Student from UA said:
Really? Or i`m full idiot or u are great scientist! =)
I have variant [tex]\int \frac{dy}{y} = \ln|y| + C = \ln |\sec(u) + \tan(u)| + C[/tex] and want to complete it, but i`m interesting in u`r variant of solve too... =)

Err... well, this maybe the result of missing so many lectures. But don't worry, you'll keep up with others soon enough. :)
In these cases, we should let t = 1 + u, and k = 1 - u ~~~> dt = du, and dk = -du, we have:

[tex]... = \frac{1}{2} \left( \int \frac{dt}{t} - \int \frac{dk}{k} \right) = \frac{1}{2} (\ln|t| - \ln|k|) + C = \frac{1}{2} \ln \left| \frac{t}{k} \right| + C = \frac{1}{2} \ln \left| \frac{1 + u}{1 - u} \right| + C[/tex] :)
 
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  • #90
VietDao29 said:
Whoops, I didn't read the whole a 6-page thread. Just skim through the main discussion :blushing: :biggrin:.
Ok, if x = 2 tan(u) ~~~> x / 2 = tan(u)
Note that [tex]u \in \left] - \frac{\pi}{2}; \ \frac{\pi}{2} \right[[/tex], so cos u > 0, we have:
[tex]\tan u = \frac{x}{2} \Rightarrow 1 + \tan ^ 2 u = 1 + \frac{x ^ 2}{4} \Rightarrow \sec ^ 2 u = 1 + \frac{x ^ 2}{4} \Rightarrow \sec u = \sqrt{1 + \frac{x ^ 2}{4}}[/tex], so plug it to the expression above, yields:
[tex]... = \ln \left| \sqrt{1 + \frac{x ^ 2}{4}} + \frac{x}{2} \right| + C[/tex]

U are really grandmaster. How old are u? tell me please =)
 
  • #91
So the last is 10 =)
tell please what i must take for u and what for dv...
 
  • #92
Student from UA said:
So the last is 10 =)
tell please what i must take for u and what for dv...

Hootenanny said:
Question 10 is just another application of the product rule. Let u=(1-3x) and dv=cos(2x)

[color="#black"]bollocks and stuff[/COLOR]
 
  • #93
Student from UA said:
So the last is 10 =)
tell please what i must take for u and what for dv...

Post #67, and post #71 should answer that:
(1 - 3x) is an algebraic function, and cos(2x) is a trig function. The one that stands before the other in the sequence LIATE should be chosen for u, and the rest for dv. So what should be u, and dv this time?
 
  • #94
[tex]\int cos2x dx[/tex] How to do?
 
  • #95
LIATE -what is it? givee me link . i want read it.
 
  • #96
Student from UA said:
[tex]\int cos2x dx[/tex] How to do?

You should let u = 2x.
When you have to deal with:
[tex]\int \sin(ax + b) dx[/tex]
[tex]\int \cos(ax + b) dx[/tex]
[tex]\int \tan(ax + b) dx[/tex]
[tex]\int e ^ {ax + b} dx[/tex]
[tex]\int \frac{1}{ax + b} dx[/tex]
...
You should let u = ax + b
Or, in general, when deadling all the function f(ax + b), where [tex]\int f(x) dx[/tex] can be found easily, make the u-substitution u = ax + b.

LIATE -what is it? givee me link . i want read it.
It's right in the post #67. :)
 
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  • #97
Liate -good. I will write in my note... =) very interesting rule
 
  • #98
U know.. I have done the 10th. It`s VICTORY! TAda! =)
 
  • #99
Thx! Thx To All WHo help ME to do my Home Work! U`re a very good persons and Very very Clever. And I don`t know why a lot of XXXXX say that USA - And Americans are very ... O don`t think so... U are very friendly people and can help other people in any time. One more thx. If i would have some qustions in future, i imidiatly will go to this nice foorum to nice people =) Thx a lot.!
 
  • #100
Well.. we're not all from the USA... in fact I don't think anyone who helped in this thread is!

Anyway, I'm glad you got your work done. Good luck in the future.
 
  • #101
don`t mater where are u from! becouse u are super mens =) i`m very happy that i`m find u!
 
  • #102
That was one of the funniest threads I've ever read through.
 
  • #103
BSMSMSTMSPHD said:
That was one of the funniest threads I've ever read through.
And why would you say that?
 
  • #104
It was like reading a really dramatic story that I didn't know the ending to. At times, I thought the poor guy wasn't going to get the answers he needed. But, he kept working hard, and eventually came through. Then he proclaimed VICTORY, and thanked the USA! It was straight out of Hollywood.

Usually I don't read through 100-post threads, but this was a next-clicker (page-turner).
 
<h2>What is an integral?</h2><p>An integral is a mathematical concept that represents the area under a curve on a graph. It is used to find the total value or quantity of something that is continuously changing.</p><h2>Why is it important to solve integrals?</h2><p>Solving integrals is important in many fields of science and engineering, as it allows us to calculate important quantities such as velocity, acceleration, and volume. It also helps us understand the behavior of a system over time.</p><h2>What are some common methods for solving integrals?</h2><p>Some common methods for solving integrals include substitution, integration by parts, and using trigonometric identities. There are also numerical methods, such as the trapezoidal rule and Simpson's rule, that can be used to approximate the value of an integral.</p><h2>What are the different types of integrals?</h2><p>The two main types of integrals are definite integrals and indefinite integrals. A definite integral has specific limits of integration and gives a numerical value, while an indefinite integral has no limits and represents a family of functions.</p><h2>How can I improve my skills in solving integrals?</h2><p>The best way to improve your skills in solving integrals is through practice. Start with simpler integrals and work your way up to more complex ones. It can also be helpful to review the basic rules and techniques for solving integrals and to seek out additional resources, such as textbooks or online tutorials.</p>

What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to find the total value or quantity of something that is continuously changing.

Why is it important to solve integrals?

Solving integrals is important in many fields of science and engineering, as it allows us to calculate important quantities such as velocity, acceleration, and volume. It also helps us understand the behavior of a system over time.

What are some common methods for solving integrals?

Some common methods for solving integrals include substitution, integration by parts, and using trigonometric identities. There are also numerical methods, such as the trapezoidal rule and Simpson's rule, that can be used to approximate the value of an integral.

What are the different types of integrals?

The two main types of integrals are definite integrals and indefinite integrals. A definite integral has specific limits of integration and gives a numerical value, while an indefinite integral has no limits and represents a family of functions.

How can I improve my skills in solving integrals?

The best way to improve your skills in solving integrals is through practice. Start with simpler integrals and work your way up to more complex ones. It can also be helpful to review the basic rules and techniques for solving integrals and to seek out additional resources, such as textbooks or online tutorials.

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