Help with IntegralS Very Important Quick

[cristo]

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 . That bandwagons getting pretty full now; I may have to think of jumping on!

 

[Student from UA]

I`ve started... and stop. 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?

 

[Student from UA]

i mean.. show integration by part becouse not has quite understood

 

[Student from UA]

And what about the 7( " ln (sec u + tan u) + C" if x=2tgu.)?

 

[VietDao29]

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:
$$\int \frac{dx}{\sqrt{x ^ 2 + \alpha ^ 2}}$$, one should think right away about making the trig-substitution $$x = \alpha \tan t , \quad t \in \left] -\frac{\pi}{2} ; \ \frac{\pi}{2} \right[$$
It goes like this:
$$x = \alpha \tan t \Rightarrow dx = (\alpha \tan t)'_t dt = \alpha \frac{1}{\cos ^ 2 t} dt$$
Your integral will become:
$$... = \int \frac{\frac{dt}{\cos ^ 2 t}}{\sqrt{\alpha ^ 2 + \alpha ^ 2 \tan ^ 2 t}} = \int \frac{dt}{ \cos ^ 2 t \sqrt{\alpha ^ 2 \left(1 + \tan ^ 2 t \right)}}$$
$$= \int \frac{dt}{\cos ^ 2 t \sqrt{\frac{1}{\cos ^ 2 t}}}$$, since $$t \in \left] -\frac{\pi}{2} ; \ \frac{\pi}{2} \right[$$, so cos t > 0, we have:
$$... = \int \frac{dt}{\cos ^ 2 t \sqrt{\frac{1}{\cos ^ 2 t}}} = \int \frac{dt}{\cos ^ 2 x \frac{1}{\cos t}} = \int \frac{dt}{\cos t}$$
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.
$$... = \int \frac{\cos t dt}{\cos ^ 2 t} = \int \frac{\cos t dt}{1 - \sin ^ 2 t}$$
Now let u = sin t, du = cos t dt, so the integral becomes:
$$... = \int \frac{du}{1 - u ^ 2} = ...$$, pretty straightforward from here. Can you go from here? :)

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

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Number 9, you should do it using $$u = \ln (x) \quad dv = x ^ {-\frac{1}{2}} dx$$. 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 $$u = \ln x x ^ {-\frac{1}{2}} \quad dv = dx$$:
$$u = \ln x x ^ {-\frac{1}{2}} \Rightarrow du = \left( (\ln(x))' x ^ {-\frac{1}{2}} + \ln (x) \left( x ^ {-\frac{1}{2}} \right) ' \right) dx$$
$$= \left( \frac{1}{x \sqrt x} - \frac{1}{2} \frac{\fbox{\ln(x)}}{x ^ {\frac{3}{2}}} \right) dx$$

 

[cristo]

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!)

 

[VietDao29]

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.

 

[Student from UA]

about 9.. u right i forgot that (uv)`= u`v+v`u

 

[VietDao29]

I`ve started... and stop. 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?

$$dv = x ^ {- \frac{1}{2}} dx \Rightarrow v = \int dv = \int x ^ {- \frac{1}{2}} dx = 2 \sqrt{x} + C = 2 \sqrt{x}$$ (C can be any constant), so we choose C = 0 to make it look more simple.
You can write $$dv = x ^ {- \frac{1}{2}} dx \Rightarrow v = 2 \sqrt{x}$$ for short. :)

 

[Student from UA]

about 7: $$... = \int \frac{du}{1 - u ^ 2} = ...$$ 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..

 

[Student from UA]

in 9th result : 2ln(x)sqrtx - 4sqrtx +c right?

Yes right... i`ve just compare with cal101 =) he he =)

 

[Student from UA]

there is 7 and 10 not comleted... agrrr

 

[VietDao29]

about 7: $$... = \int \frac{du}{1 - u ^ 2} = ...$$ 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..

$$... = \int \frac{du}{(1 - u) (1 + u)} = \frac{1}{2} \int \frac{(1 - u) + (1 + u)}{(1 - u) (1 + u)} du = \frac{1}{2} \int \left( \frac{1 - u}{(1 - u) (1 + u)} + \frac{1 + u}{(1 - u) (1 + u)} \right) du$$

$$= \frac{1}{2} \int \left( \frac{1}{1 + u} + \frac{1}{1 - u} \right) du = ...$$ 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:
$$\int \sec(x) dx = \int \sec(x) \frac{\sec(x) + \tan (x)}{\sec(x) + \tan (x)} dx = \int \frac{\sec ^ 2 x + \sec (x) \tan (x)}{\sec(x) + \tan (x)} dx$$
Let u = sec(x) + tan(x)
$$\Rightarrow du = \left( \frac{1}{\cos x} + \tan(x) \right)'_x dx = \left( \frac{\sin (x)}{\cos ^ 2 (x)} + \frac{1}{\cos ^ 2 x} \right) dx = \left( \sec(x) \tan(x) + \sec ^ 2 (x) \right) dx$$
The integral will become:
$$\int \frac{du}{u} = \ln|u| + C = \ln |\sec(x) + \tan(x)| + C$$

 

[Student from UA]

VietDao29 u are the monster of Integrals! My teacher is stupid if her compare with u.

 

[Student from UA]

$$... = \int \frac{du}{(1 - u) (1 + u)} = \frac{1}{2} \int \frac{(1 - u) + (1 + u)}{(1 - u) (1 + u)} du = \frac{1}{2} \int \left( \frac{1 - u}{(1 - u) (1 + u)} + \frac{1 + u}{(1 - u) (1 + u)} \right) du$$

$$= \frac{1}{2} \int \left( \frac{1}{1 + u} + \frac{1}{1 - u} \right) du = ...$$ 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:
$$\int \sec(x) dx = \int \sec(x) \frac{\sec(x) + \tan (x)}{\sec(x) + \tan (x)} dx = \int \frac{\sec ^ 2 x + \sec (x) \tan (x)}{\sec(x) + \tan (x)} dx$$
Let u = sec(x) + tan(x)
$$\Rightarrow du = \left( \frac{1}{\cos x} + \tan(x) \right)'_x dx = \left( \frac{\sin (x)}{\cos ^ 2 (x)} + \frac{1}{\cos ^ 2 x} \right) dx = \left( \sec(x) \tan(x) + \sec ^ 2 (x) \right) dx$$
The integral will become:
$$\int \frac{du}{u} = \ln|u| + C = \ln |\sec(x) + \tan(x)| + C$$

Agrrr See post 63! =)$$\int \frac{du}{u} = \ln|u| + C = \ln |\sec(x) + \tan(x)| + C$$ I have the same but how to solve "$$\int \frac{dy}{y} = \ln|y| + C = \ln |\sec(u) + \tan(u)| + C$$" if x=2tgu

 

[Student from UA]

$$= \frac{1}{2} \int \left( \frac{1}{1 + u} + \frac{1}{1 - u} \right) du = ...$$ You should be able to go from here. :)

Really? Or i`m full idiot or u are great scientist! =)
I have variant $$\int \frac{dy}{y} = \ln|y| + C = \ln |\sec(u) + \tan(u)| + C$$ and want to complete it, but i`m interesting in u`r variant of solve too... =)

 

[VietDao29]

Agrrr See post 63! =)$$\int \frac{du}{u} = \ln|u| + C = \ln |\sec(x) + \tan(x)| + C$$ I have the same but how to solve "$$\int \frac{dy}{y} = \ln|y| + C = \ln |\sec(u) + \tan(u)| + C$$" if x=2tgu

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

 

[Hootenanny]

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

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...

 

[VietDao29]

Really? Or i`m full idiot or u are great scientist! =)
I have variant $$\int \frac{dy}{y} = \ln|y| + C = \ln |\sec(u) + \tan(u)| + C$$ 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:

$$... = \frac{1}{2} \left( \int \frac{dt}{t} - \int \frac{dk}{k} \right) = \frac{1}{2} (\ln|t| - \ln|k|) + C = \frac{1}{2} \ln \left| \frac{t}{k} \right| + C = \frac{1}{2} \ln \left| \frac{1 + u}{1 - u} \right| + C$$ :)

 

[Student from UA]

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

U are really grandmaster. How old are u? tell me please =)

 

[Student from UA]

So the last is 10 =)
tell please what i must take for u and what for dv...

 

[Hootenanny]

So the last is 10 =)
tell please what i must take for u and what for dv...

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

[color="#black"]bollocks and stuff[/COLOR]

 

[VietDao29]

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?

 

[Student from UA]

$$\int cos2x dx$$ How to do?

 

[Student from UA]

LIATE -what is it? givee me link . i want read it.

 

[VietDao29]

$$\int cos2x dx$$ How to do?

You should let u = 2x.
When you have to deal with:
$$\int \sin(ax + b) dx$$
$$\int \cos(ax + b) dx$$
$$\int \tan(ax + b) dx$$
$$\int e ^ {ax + b} dx$$
$$\int \frac{1}{ax + b} dx$$
...
You should let u = ax + b
Or, in general, when deadling all the function f(ax + b), where $$\int f(x) dx$$ 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. :)

 

[Student from UA]

Liate -good. I will write in my note... =) very interesting rule

 

[Student from UA]

U know.. I have done the 10th. It`s VICTORY! TAda! =)

 

[Student from UA]

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.!

 

[cristo]

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.

 

[Student from UA]

don`t mater where are u from! becouse u are super mens =) i`m very happy that i`m find u!

 

[BSMSMSTMSPHD]

That was one of the funniest threads I've ever read through.

 

[Hootenanny]

That was one of the funniest threads I've ever read through.

And why would you say that?

 

[BSMSMSTMSPHD]

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).