Help with IntegralS Very Important Quick

[Student from UA]

maybe some people have password from cal101.com? Please help... Becouse it can show integration step by step

 

[Hootenanny]

Okay, have you read the link I gave you in post #55?

 

[Student from UA]

ok. explain me how solve " ln |sec u + tan u| + C" if x=2tgu

 

[Student from UA]

U know . I`ve done 3 and 4. =)
See in atach and say right or wrong.
In 4 a had two variants of result. Please tell which off them is right.

 

[Student from UA]

where are u ?! Please don`t leave me alone =(

 

[Student from UA]

i`ve done 9th. But I`m comapare it with cal101.com result...
Hm we have different. SO please where tell where a ihave mistake.
Look atach

 

[VietDao29]

You did number 3, and 4 correctly. Congratulations. :) In number 4, you can leave it in the form:
\frac{2}{\ln 2} 2 ^ {\frac{x}{2}} + C, it's okay.

Number 5 is a little bit messy.
When doing Integration by Parts, you should remember the word LIATE. It stands for Logarithmic, Inverse Trig, Algebraic, Trig, Exponential. That is the order to go to pick your "u".
So in your problem, ln(x) is a logarithmic function, and x ^ {- \frac{1}{2}} is algebraic, and u of course will be ln(x), since logarithmic stands before algebraic. The rest should be dv:
\int \frac{\ln x}{\sqrt{x}} dx = \int \ln (x) x ^ {-\frac{1}{2}} dx, your u, and dv will be:
u = \ln (x) \quad dv = x ^ {-\frac{1}{2}} dx
So what should your du, and v be? Can you go from here? :)

P.S: And you should also note that "dx" is often put at the end of the expression. Say x dx, or x²dx, you should not write dx x, or dx x².

 

[Student from UA]

with 10... i don`t know how to start! Please people help.
Remind :
7 - not competed. explain me how solve " ln (sec u + tan u) + C" if x=2tgu.
9 - please where tell where a I have mistake.
10 -Try to solve it... or tell how to start...
And that's all! =) My home work will be completed... Really please help, becouse now 20.25 pm on my clock... and i must go to bed... becouse i`m early get up, nearly 6 o`clock. (university is far from me)

 

[Student from UA]

ok.. ill try to make 9th =) u sad 5 but it`s 9th =)

 

[Hootenanny]

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

With respect to question 9, I'm afraid I can't understand your writing; but if you follow VietDao29's instructions, you should obtain the correct answer.

 

[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 =)