two calc 2 problems

[EZ-Bar]

First one I thought was right...guess not. Can you tell me whats wrong?

Integral of (e^3x) / (e^3x) + 1 * dx

u = e^3x + 1
du = e^3xdx

integral of 1 /u * du

= ln|u|
= ln|e^3x + 1|

right?


next problem I have no fockin clue

Integral of (Sin(41x)) / 1 + (cos(41x))^2


will rep of course

[MathGuy]

Originally Posted by EZ-Bar

First one I thought was right...guess not. Can you tell me whats wrong?

Integral of (e^3x) / (e^3x) + 1 * dx

u = e^3x + 1
du = e^3xdx

integral of 1 /u * du

= ln|u|
= ln|e^3x + 1|

right?


next problem I have no fockin clue

Integral of (Sin(41x)) / 1 + (cos(41x))^2


will rep of course

Your integral of e^3x + 1 is wrong.

sinx and cosx are related (ie. one is the integral of the other).

[ex3e1989]

Originally Posted by EZ-Bar

First one I thought was right...guess not. Can you tell me whats wrong?

Integral of (e^3x) / (e^3x) + 1 * dx

u = e^3x + 1
du = e^3xdx

integral of 1 /u * du

= ln|u|
= ln|e^3x + 1|

right?


next problem I have no fockin clue

Integral of (Sin(41x)) / 1 + (cos(41x))^2


will rep of course

u = e^3x + 1
du = 3e^3xdx
u forgot the 3 brah

so your final answer would be
= (1/3)ln|e^3x + 1|

[alexk7]

Originally Posted by EZ-Bar

u = e^3x + 1
du = 3e^3x

i have no clue about what the rest means your notation is awful

[EZ-Bar]

Originally Posted by ex3e1989

u = e^3x + 1
du = 3e^3xdx
u forgot the 3 brah

so your final answer would be
= (1/3)ln|e^3x + 1|

ahh wow lol thanks

Originally Posted by alexk7

i have no clue about what the rest means your notation is awful

lol

alright how about this

Integral of: Sin(41x) divided by 1 + (cos(41x))squared

right now I just got

arctan times cos(41x) all divided by -sin(41x)

[ex3e1989]

Originally Posted by EZ-Bar

ahh wow lol thanks



lol

alright how about this

Integral of: Sin(41x) divided by 1 + (cos(41x))squared

right now I just got

arctan times cos(41x) all divided by -sin(41x)

howd u get that

[EZ-Bar]

uhh lol i am probably wrong but heres what I did

I did

u = cos(41x)
du = -sin(41x)dx
du/-sin(41x) = dx

so 1-/sin(41x) integral 1/(1+u^2 )

so 1/-sin(41x) * arctan(u)

so arctan(cos(41x)) all divided by -sin(41x)

[ex3e1989]

let u = cos(41x)
du/dx = -41 * sin(41x)
(-1/41).(du/dx) = sin(41x) (see what i did there?)
so it becomes (-1/41) integral 1/(1+u^2) du (replace the sin(41x) instead of the dx like u did above)
= (-1/41)arctan(u)
=(-1/41)arctan(cos41x)

[ex3e1989]

in ur one u have done 1-/sin(41x) integral 1/(1+u^2 )
which is saying that 1-/sin(41x) is a constant, when it is not.

[EZ-Bar]

ahhhh ok that makes sense

can you elaborate on how you could pull the -41 out in the derivative?

[ex3e1989]

it's the rule brah

d(sin(Ax))/dx = Acos(Ax)

likewise

d(cos(Ax))/dx = -Asin(Ax)

it's just like d(e^ax)/dx = ae^ax

you can prove it using the chain rule