Help with IntegralS Very Important Quick

[Student from UA]

so the rsult of 5: $$\intdu=\arcsin \frac{x}{2} + C$$ ?

 

[Gib Z]

I have no idea what the "tg" is, but I deduce its the tangent function.

Even if it isnt, let $$u=\sqrt{x}$$ Then $$\frac{du}{dx} = \frac{1}{2u}, dx = 2u du$$

so the integral becomes

$$\int \tg u \frac{2u}{u} du = 2\int \tg (u) du$$

EDIT: As to post 36, q5, yes that is the answer. The reason you can not write it as you did previously is because it is only valid for n E Z, but we need it to be valid for all real values of n.

 

[Student from UA]

Gibs $$u=\sqrt{x}$$
$$du=\frac{1}{2\sqrt{x}}dx$$ then $$dx=2\sqrt{x}du$$

 

[Gib Z]

Check your typing, you forgot to use the right hand brace } instead of right breacket ). Anyway, then you are correct. However, didn't we say $u=\sqrt{x}$?

 

[Gib Z]

You know either way it doesn't matter. You are stuck with
$$2\int \frac{\sin x}{\cos x} dx$$ which can be solved by letting u = cos x

 

[Student from UA]

right result?

 

[Gib Z]

Nope.

$$2\int \frac{\sin x}{\cos x} dx$$
let u = cos x, then du = - sin x dx
$$-2\int \frac{1}{u} du = -2\log_e u + C = -2 \log_e (\cos x) + C$$

 

[Student from UA]

Nope.

$$2\int \frac{\sin x}{\cos x} dx$$
let u = cos x, then du = - sin x dx
$$-2\int \frac{1}{u} du = -2\log_e u + C = -2 \log_e (\cos x) + C$$

U right, i agree with u. I made a child mistake =(
simply I sit near computer and do homework already 6 hours and have square head

 

[Student from UA]

start the 7th. Stop in the middle. See in atach. :zzz:

 

[cristo]

So you have the integral of sec u= 1/cos u. To compute this, try multiplying top and bottom by sec u+tan u.

 

[Student from UA]

i don`t understand u cristo =(

 

[Student from UA]

stop don`t tell me anything 5 minut i try tio solve it again.

 

[Student from UA]

secu^2= 1+tgu^2 => secu=sqrt(1+tgu^2)
so compare 3rd step and last in atach . hmm

 

[cristo]

$$\sec u=\sec u\cdot\left(\frac{\sec u+\tan u}{\sec u+\tan u}\right)$$. Can you integrate this?

[Hint: How is the numerator related to the denominator?]

 

[Student from UA]

ou... sorry =) how to delete... -) it`s mistake

 

[Student from UA]

to cristo... no i can`t =*(

 

[cristo]

Ok.. expand tp give $$\sec u=\sec u\cdot\left(\frac{\sec u+\tan u}{\sec u+\tan u}\right)=\frac{\sec^2u+\tan u\sec u}{\sec u+\tan u}.$$ You worked out that d/du(tan u)=sec^2u. It can also be shown that d/du(sec u)=tan u.sec u, and thus the numerator is the exact derivative of the denominator.

Do you know how to integrate a function of the form $$\frac{f'(u)}{f(u)}$$? [Hint: Think of the most simple function of this form; namely 1/x].

 

[Student from UA]

half of your text i can`t understand. and the f`(u)/f(u) i don`t know maybe =( and i `m really tied =(

 

[Student from UA]

and id look to all that u sad ... i can`t stand =( really. can u write all step by step to result? maybe if i look i can stand

 

[Hootenanny]

If you're really tired, you should take a break; you're going to gain very little from working when your exhausted. With respect to cristo's suggestion a full proof can be found at http://math2.org/math/integrals/more/sec.htm .

 

[Student from UA]

Hootenanny: all were well if not one but. Tommorow i need to go to university with complete homework... and now i haven`t free time for relax =( head is been ill but i try to solve this stupid integrals... And it very hard to me (little ukrainian boy) study integrals on english and speak with u becouse my English not well...maybe bad.

 

[Student from UA]

ur tan it`s my tg? tangens? or what?

 

[Student from UA]

tan=sin/cos? if yes - tan it`s tg

 

[Hootenanny]

Still, not handing in homework on time isn't the end of the world and if you have been ill, I'm sure you can apply for special circumstances. However, I understand that it must be difficult for you to communicate on a predominantly english speaking forum; at least its mathematics homework your doing at not english/ukrainian

And yes, tan is our tangent (your tg) function.

 

[Student from UA]

hm... I am my situationi it`s tne end of world, becouse my my scholarship depends on it. It`s really hard to understand.. =(
Their does not interest, was ill I or no. Simply, it is necessary to give homework in time.
Please help me =( U`re my last chans half homework we do...

 

[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.