General

[quote=NoobCake]You put more work into that than I have all semester.[/quote]
Thanks, my goal in life is to become a teacher.

2) Indefinite integral of 2dx/(xln(5x))
a) Rewrite the problem.
= Integral(2/(xln(5x)) [I omit most dx/du/dv's for sake of less typing]
b) Pull out the constant.
= 2 * Integral(1/(xln(5x))
c) Use integral substitution (see formula of my answer on question #1b if you forget).
[*]u = 5x
[*]du = 5dx
[*]dx = (1/5)du
= 2(Integral( (1/(xln(u)) * (1/5) )
= 2(Integral((1/(5xln(u)))
[*]By basic algebra x = u/5, because u = 5x
= 2(Integral(1 /(5uln(u)) /5 )
d) Pull the constant out of this mess.
= 2/(5/5) (Integral(1/(uln(u)))
= 2 (Integral(1/(uln(u)))
e) Apply integral substitution again.
v = ln(u)
dv = (1/u) du
du = udv
= 2(Integral(1/(uv) udv)
= 2(Integral(1/(v))
f) Recall that the integral of 1/x = ln(x).
= 2ln(v)
g) Substitute v back into u.
[*]v = ln(u)
= 2ln(ln(u))
h) Substitute u back into x.
[*]u = 5x
= 2(ln(ln(5x))
[*]Add a constant.
[b]= 2(ln(ln(5x)) + C[/b]

3) Definite integral of (7+cos^(3)u)/(cos^(2)u) from 0 to pi/4
a) Find the indefinite integral.
Integral((7+cos^(3)u)/(cos^(2)u))
b) Use the addition rule of integrals. Integral(f(x) + g(x)) = Integral(f(x)) + Integral(g(x))
= Integral(7/(cos^(2)u)) + Integral(cos^(3)u /cos^(2)u)
c) For the first integral, Integral(7/(cos^(2)u)):
= 7 * Integral(1/cos^(2)u)
[*]Recall that Integral(1/cos^(2)u) = tan(u)
= 7tan(u)
d) For the second integral, Integral(cos^(3)u /cos^(2)u):
Integral(cos^(3)u /cos^(2)u) = Integral(cos(u))
[*]Recall that Integral(cos(u)) = sin(u)
= sin(u)
e) Add both computed integrals back together.
= 7tan(u) + sin(u)
[b]= 7tan(u) + sin(u) + C[/b]

All done, and hopefully all correct.

[quote=Yumtoast]1) Definite integral of dx/(4x+6) from 0 to 2
a) Rewrite the problem.
[*]My notation for integrals is Integral(x0, x1, f(x)) (read as integral of f(x) from x0 to x1).
= Integral(0, 2, 1/(4x+6))
b) Find the indefinite integral via substitution.
[*]Note that Integral(f(g(x)) * g(x)dx) = Integral(f(u)du) with u = g(x).
[*]u = 4x +6
[*]du = 4dx
[*]dx = 1/4 du
= Integral((1/u)(1/4))
c) Pull the constant out of the integral.
= (1/4) * Integral(1/u)
d) Use the integral rule, Integral(1/u) = ln(u)
= (1/4) * ln(u)
e) Substitute u back into the function.
[*]u = 4x + 6
= (1/4) * ln(4x + 6)
= ln(4x + 6)/4
f) Compute the definite integral at x = 2.
= ln(4x + 6)/4
= ln((4 * 2) + 6)/4
= ln(14)/4
g) Compute the definite integral at x = 0.
= ln(4x + 6)/4
= ln((4 * 0) + 6)/4
= ln(6)/4
h) Subtract the two answers.
[*]Recall that Integral(x0, x1, f(x)) = F(x1) - F(x0).
[*]F(x1) = ln(14)/4
[*]F(x0) = ln(6)/4
= (ln(14)/4) - (ln(6)/4)
[*]Simplify
[b]= (1/4)ln(7/3)[/b]

I'll answer the other problems later.[/quote]

You put more work into that than I have all semester.

1) Definite integral of dx/(4x+6) from 0 to 2
a) Rewrite the problem.
[*]My notation for integrals is Integral(x0, x1, f(x)) (read as integral of f(x) from x0 to x1).
= Integral(0, 2, 1/(4x+6))
b) Find the indefinite integral via substitution.
[*]Note that Integral(f(g(x)) * g(x)dx) = Integral(f(u)du) with u = g(x).
[*]u = 4x +6
[*]du = 4dx
[*]dx = 1/4 du
= Integral((1/u)(1/4))
c) Pull the constant out of the integral.
= (1/4) * Integral(1/u)
d) Use the integral rule, Integral(1/u) = ln(u)
= (1/4) * ln(u)
e) Substitute u back into the function.
[*]u = 4x + 6
= (1/4) * ln(4x + 6)
= ln(4x + 6)/4
f) Compute the definite integral at x = 2.
= ln(4x + 6)/4
= ln((4 * 2) + 6)/4
= ln(14)/4
g) Compute the definite integral at x = 0.
= ln(4x + 6)/4
= ln((4 * 0) + 6)/4
= ln(6)/4
h) Subtract the two answers.
[*]Recall that Integral(x0, x1, f(x)) = F(x1) - F(x0).
[*]F(x1) = ln(14)/4
[*]F(x0) = ln(6)/4
= (ln(14)/4) - (ln(6)/4)
[*]Simplify
[b]= (1/4)ln(7/3)[/b]

I'll answer the other problems later.

[quote=Mikin]Ohhh it's because ln(5) is 0. Got it! Thanks so much.[/quote]

ln(5) isn't 0... ln(1) is 0.
The problem I made was the chain rule when differentiating it.

du = 1*5dx/5x = dx/x

[quote=NoobCake]Oops my bad about #2, I just based it off looking and now I actually wrote it down...here's what I got:

u = ln(5x)
du = dx/x

2 integral du/u
= 2ln(u)
= 2ln(ln(5x)) [b]+ c[/b][/quote]

Ohhh it's because ln(5) is 0. Got it! Thanks so much.

[quote=Mikin]Okay i got number 1, but for number 2 put 10(ln(ln(5x))) but its still wrong =[[/quote]

Oops my bad about #2, I just based it off looking and now I actually wrote it down...here's what I got:

u = ln(5x)
du = dx/x

2 integral du/u
= 2ln(u)
= 2ln(ln(5x)) [b]+ c[/b]

[quote=NoobCake]1) Set u = 4x+6, du = 4dx -> du/4 = dx.
(1/4) integral du/u
= (1/4)ln(4x+6) now evaluate at 0 & 2.

2) Let u = ln(5x), du = dx/5x -> 5du = dx/x
Solve.[/quote]

Okay i got number 1, but for number 2 put 10(ln(ln(5x))) but its still wrong =[

[quote=Mikin]For the first problem is the integral ln(4x+6) and i just do ln(2+6)-ln(0)?
For the second problem I used substitution and got 10ln(ln(5x)) but it keeps saying I'm incorrect [/quote]

1) Set u = 4x+6, du = 4dx -> du/4 = dx.
(1/4) integral du/u
= (1/4)ln(4x+6) now evaluate at 0 & 2.

2) Let u = ln(5x), du = dx/5x -> 5du = dx/x
Solve.

[quote=NoobCake]1) U substitution.
2) U substitution.
3) Integration by parts.[/quote]

For the first problem is the integral ln(4x+6) and i just do ln(2+6)-ln(0)?
For the second problem I used substitution and got 10ln(ln(5x)) but it keeps saying I'm incorrect

1) U substitution.
2) U substitution.
3) Integration by parts.