@Countolaf: HELP!

[Fox Mc Cloud]

Thise are excersises with differentialequations. I don't now how to solve it and I have to know it for my exams, so maybe you can help me. If you find them, show how you did it. It's with partial integration (integral of u*dv=u*v-integral of v*du).

You have to calculae the integral of those things:

y=cos(x)*e^-x
Answer should be: 1/2 [e^-x*sin(x)-e^-x*cos(x)]+C

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y=bgcos(4x)
Answer should be: x*bgcos(4x)-1/4*(1-16x²)^1/2+C

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y=[ln(3t)]/t²
Answer should be: [-ln(3t)]/t - 1/t + C

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y=sin(t)*ln[1+sin(t)]
Answer should be: -cos(t)*ln[1+sin(t)]+t+cos(t)+C



If you can help me, then I have a surprise for you .

[Countolaf]

I haven't studied Calculus II yet, just I, so it is not very easy to me to solve these things.[a href="http://[del:countolaf]"]http://[del:countolaf][/a]

[Fox Mc Cloud]

I don't understand it, and it's main part of my exam.

[Countolaf]

Let me take an easier example I know how to answer:
If we are trying to find the value of (symbol of integral)xcos(x)dx, then we can call:

u = x
Then, du = dx
And call all the rest dv
Then, dv = cos(x)dx
Then, v = sin(x)

Now we do: (symbol of integral)udv = uv - (symbol of integral)vdu
And, using the values for u, v, du and dv we found:
(symbol of integral)xcos(x)dx = xsin(x) - (symbol of integral)sin(x)dx
And, finally:
(symbol of integral)xcos(x)dx = xsin(x) + cos(x) + C[a href="http://[del:countolaf]"]http://[del:countolaf][/a]

[Fox Mc Cloud]

Yea that's it, but now try to solve my things with this rule. I have no idea what to call u and what dv. You don't need any other rules except this one. If you can't solve it, nobody can .

[Fox Mc Cloud]

NIIIIIIIIIIIICE

3 left .

[Countolaf]

I need a time to think about the others o.o
I will go to Kumon now, I will be back in 3 hours I think.
Hope you understood my explanation on how to solve first problem

[Fox Mc Cloud]

Thx for the help .

[Countolaf]

No problem!
I hope it wasn't too colourful. o.O

[Fox Mc Cloud]

No it was good with the colors. Easy to see what you where doing. ;D

[Countolaf]

What is bgcos?
Belgian cosines? [a href="http://[del:countolaf]"]http://[del:countolaf][/a]

[Fox Mc Cloud]

bgcos=arccos=cos^-1

[Countolaf]

Didn't know arccos = bgcos
I will try to solve it now then

[Countolaf]

If we have ç bgcos(4x) dx, we can call:
u = 4x
Then, du = 4 dx

Then, ç bgcos(4x) dx = 1/4 ç bgcos(u) du .... (1)

We have that (not considering constant C):
ç bgcos(u) du = u bgcos(u) - (1 - u^2)^1/2 .... (2)

Using equation (2) in equation (1):
ç bgcos(4x) dx = 1/4
ç bgcos(4x) dx = 1/4 u bgcos(u) - 1/4 (1 - u^2)^1/2

Since u = 4x:
ç bgcos(4x) dx = x bgcos(4x) - 1/4 (1 - 16x^2)^1/2

Then, considering the constant C:
ç bgcos(4x) dx = x bgcos(4x) - 1/4 (1 - 16x^2)^1/2 + C

[Fox Mc Cloud]

Thx for solving. I think bgcos is the new notation of arccos, at least it is in europe.

[Countolaf]

1) If we have (symbol of integral) cos(x) e^-x dx, we can call (not considering constants):
u = cos(x)
Then, du = - sin(x) dx
dv = e^-x dx
Then, v = - e^-x
We can say that (symbol of integral)udv = uv - (symbol of integral)vdu
Then:
(symbol of integral) cos(x) e^-x dx = - cos(x) e^-x - (symbol of integral) e^-x sin(x) dx .... (1)

If we have (symbol of integral) e^-x sin(x) dx, we can call (not considering constants):
u = sin(x)
Then, du = cos(x) dx
dv = e^-x dx
Then, v = - e^-x
We can say that (symbol of integral)udv = uv - (symbol of integral)vdu
Then:
(symbol of integral) sin(x) e^-x dx = - sin(x) e^-x + (symbol of integral) cos(x) e^-x dx .... (2)

Using equation (2) in equation (1), we have:
(symbol of integral) cos(x) e^-x dx = - cos(x) e^-x - [- sin(x) e^-x + (symbol of integral) e^-x cos(x) dx]
(symbol of integral) cos(x) e^-x dx = - cos(x) e^-x + sin(x) e^-x - (symbol of integral) cos(x) e^-x dx
2 (symbol of integral) cos(x) e^-x dx = - cos(x) e^-x + sin(x) e^-x

Then, considering constant C:
(symbol of integral) cos(x) e^-x dx = 1/2 [sin(x) e^-x - cos(x) e^-x] + C

- Rewritten because I have corrected some errors

[Countolaf]

If we have (symbol of integral) [ln(3t)]/t² dt, we can call (not considering constant C):
u = ln(3t)
Then, du = 1/t dt
dv = 1/t² dt
Then, v = - 1/t
We can say that (symbol of integral)udv = uv - (symbol of integral)vdu
Then:
(symbol of integral) [ln(3t)]/t² dt = - ln(3t)/t + (symbol of integral) 1/t² dt
(symbol of integral) [ln(3t)]/t² dt = - ln(3t)/t - 1/t

Considering constant C:
(symbol of integral) [ln(3t)]/t² dt = - ln(3t)/t - 1/t + C

[Fox Mc Cloud]

Thanks . Maybe we should make a homework board .

[Countolaf]

(symbol of integral) sin(t) ln[1+sin(t)] dt
We can call:
u = ln[1 + sin(t)]
Then, du = cos(t)/[1 + sin(t)] (using chain rule)
dv = sin(t)
Then, v = -cos(t)

We can say that (symbol of integral) u dv = uv - (symbol of integral) v du

Then:
(symbol of integral) sin(t) ln[1+sin(t)] dt = - ln[1 + sin(t)] cos(t) + (symbol of integral) cos²(t)/[1 + sin(t)] dt


(symbol of integral) cos²(t)/[1 + sin(t)] dt
= (symbol of integral) [cos²(t)][1 - sin(t)]/[1 + sin(t)][1 - sin(t)] dt
= (symbol of integral) [cos²(t)][1 - sin(t)]/[1 - sin²(t)] dt
= (symbol of integral) [cos²(t)][1 - sin(t)]/[cos²(t)] dt
= (symbol of integral) [1 - sin(t)] dt
= cos(t) + t

Then:
(symbol of integral) sin(t) ln[1+sin(t)] dt = - ln[1 + sin(t)] cos(t) + cos(t) + t
(symbol of integral) sin(t) ln[1+sin(t)] dt = - cos(t) ln[1 + sin(t)] + t + cos(t)

Considering constant of integration C:
(symbol of integral) sin(t) ln[1+sin(t)] dt = - cos(t) ln[1 + sin(t)] + t + cos(t) + C

[Fox Mc Cloud]

Thanks Countolaf, you're great in maths .