Calculus Help Needed!!!

[Iilac]

Hi

I have some intergration problems that are simply beyond me and any help would be appreciated.

1)Intergration of x [arcsin(x^2)] dx

2)Intergration of ln(x^4)/x dx

3)Intergration of dx/(1+2e^3x)

thanks

 

[Lager]

Hopefully somebody helps you.

WHY CALCULUS ON FRIDAY NIGHT?

 

[TechnoKid]

something about integration by parts pops up to my mind...?bah..its a friday and i dont really want to look at my notes for my calc test on monday....

 

[Muzzan]

ln(x^4) / x = 4 * ln(x) * 1/x.

1/x is the derivative of ln(x), so the integral is of the form k * f(x) * f'(x), which looks a lot like something you'd get from the chain rule. Assume the integral is f(g(x)) for some suitable functions f and g (and let's forget about all constants for a while). The derivative of the integral is f'(g(x)) * g'(x), but this was supposed to be equal to ln(x) * 1/x. Obviously g'(x) = 1/x, which means g(x) = ln(x). So:

f'(ln(x)) * 1/x = ln(x) * 1/x
<=>
f'(ln(x)) = ln(x)

Obviously f'(x) = x will work, but this gives f(x) = x^2/2. So the integral looks something like (ln(x))^2 / 2 + C. Differentiating this, we get ln(x) * 1/x, so we have to multiply everything by 4 to get what we wanted. The final answer is 2ln(x)^2 + C.

 

[Iilac]

Thanks Muzzan for your help. This stuff is due on monday, so I wanted to get a head start on it.

 

[Iilac]

Bump for calculus help

 

[Iilac]

Bump

 

[Darien]

for the first one, set u = 2x. then it's just integrating the arcsin function.

the 2nd one was posted.

the last one, i'm not sure how to formally derive it. since the derivative of lnu is u'/u, try and find a function that'll give you 1/u. (eg: something - lnu)

 

[Iilac]

Thank you Darien. Anybody else have more to add?

 

[Iilac]

bump for calculus

 

[Iilac]

what is the integration of arcsin?

 

[waffel]

ok, integral of dx/(1 + 2e^3x)

let u = e^3x, so du = 3e^3x dx = 3udx
so dx = du/(3u)

substitute back in, you get the integral of du/(3u*(1 + 2u))

now there's an integral for this, integral of du/(u*(a + bu)) = (1/a)*ln(u/(a + bu)) + C
here, a = 1 and b = 2, and the whole thing is multiplied by 1/3

so you get (1/3)*ln(u/(1 + 2u)) + C
plug e^3x in for u, and the result is (1/3)*ln(e^3x/(1 + 2e^3x)) + C

just that easy!

 

[waffel]

Originally posted by: Iilac
what is the integration of arcsin?

integration of arcsin(u) du = u*arcsin(u) + sqrt(1 - u^2) + C

 

[Iilac]

Thank you waffel What is that integration called?

 

[Iilac]

Here is what I got for problem 1

1/2*1/(square root of 1-(x^2)^2) + C. Is that right?

 

[KnickNut3]

Anyone know how to create a full Fourier series of cos(x/2) - sin(x) ? When I try to solve for a sub n I get cos(x/2)cos(nx) - sin(x)cos(nx). Definitely non-integrable... anyone know?

 

[Iilac]

How about this one

The integration of sin ^2(2x)*cos (x) dx

 

[Iilac]

Bump for calc help

 

[Iilac]

bump