Integral please check

prime-factor · Jan 20, 2009

1. The problem statement, all variables and given/known data

∫ -60x³tan(3x⁴)dx

2. Relevant equations

integration rules

3. The attempt at a solution

=> ∫ -60x³tan(3x⁴)dx

=> u = (3x⁴)

=> du = 12x³dx

=> (-5)∫ (1/5)-60x³tan(3x⁴)dx

=> -5 ∫ tan u

=> -5 .-ln cos(u) , where u = (3x⁴)

=> 5 . ln cos (3x⁴)

I am using this integral to check an original equation which I differentiated which was:
ln (cos⁵(3x⁴) , so where have I gone wrong when integrating, because I'm not getting it back ? ( I have started just learning the basics of integration so be nice :) )

Mark44 · Jan 21, 2009

prime-factor said: ↑

1. The problem statement, all variables and given/known data

∫ -60x³tan(3x⁴)dx

2. Relevant equations

integration rules

3. The attempt at a solution

=> ∫ -60x³tan(3x⁴)dx

=> u = (3x⁴)

=> du = 12x³dx

=> (-5)∫ (1/5)-60x³tan(3x⁴)dx

=> -5 ∫ tan u

=> -5 .-ln cos(u) , where u = (3x⁴)

=> 5 . ln cos (3x⁴)

I am using this integral to check an original equation which I differentiated which was:
ln (cos⁵(3x⁴) , so where have I gone wrong when integrating, because I'm not getting it back ? ( I have started just learning the basics of integration so be nice :) )

I don't see anything wrong with what you have, short of a missing constant of integration and missing absolute values around cos(u) and a few other minor things.

I suspect that you made an error in your differentation and got the result you did.

Assuming you started with ln (cos⁵(3x⁴)),
d/dx[ ln (cos⁵(3x⁴)]
= 1/cos⁵(3x⁴) * d/dx [cos⁵(3x⁴]
= 1/cos⁵(3x⁴) * (5)*cos⁴(3x⁴) * d/dx( cos(3x⁴))
= 1/cos⁵(3x⁴) * (5)*cos⁴(3x⁴) * -sin(3x⁴) * d/dx (3x⁴)
= 1/cos⁵(3x⁴) * (5)*cos⁴(3x⁴) * -sin(3x⁴) * 12x³)
[tex]= \frac{-60 x^3 sin(3x^4)}{cos(3x^4}[/tex]
= -60x³ tan(3x⁴)

One minor thing I referred to is your misuse of the "implies" symbol where you should be using "equals." Each of your integration steps involves an expression, not a statement, so from one step to the next you should use "=".

You used ==> correctly in your substitution, namely

u = (3x⁴)
=> du = 12x³dx
Here, the first line is an equation, a kind of statement, and the second equation follows from it.

prime-factor · Jan 21, 2009

Thanks for the thorough response :).

But I am still not sure why I can't get back: 'ln (cos⁵(3x4)'

I differentiated it as you did and got the same answer, but
I am not getting it back when integrating?

Mark44 · Jan 21, 2009

When you say "get back" what are you starting with?

prime-factor · Jan 21, 2009

Okay. Sorry I explained it so badly.

I started with a function which was:

y = ln cos⁵(3x⁴)

differentiating this gives: -60x³tan(3x⁴)

Now why is it that when I integrate '-60x³tan(3x⁴)' I don't get back the original function which was 'y = ln cos⁵(3x⁴)'.

Mark44 · Jan 21, 2009

You sort of do. ln w⁵ = 5 ln w.

prime-factor · Jan 21, 2009

Oh My God... I can't believe I did not see that before!!

I forgot that cos⁵(3x⁴) = (cos(3x⁴)⁵

and hence with the ln in front it is treated like a basic operation.

You are very kind for helping me with this and being so pacient :).

Thanks Again.

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Integral please check

prime-factor

Mark44

Staff: Mentor

prime-factor

Mark44

Staff: Mentor

prime-factor

Mark44

Staff: Mentor

prime-factor

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