1. Jan 20, 2009

### prime-factor

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 :) )

Last edited: Jan 20, 2009
2. Jan 21, 2009

### Staff: Mentor

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)
$$= \frac{-60 x^3 sin(3x^4)}{cos(3x^4}$$
= -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.

3. Jan 21, 2009

### prime-factor

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?

4. Jan 21, 2009

### Staff: Mentor

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

5. Jan 21, 2009

### prime-factor

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)'.

6. Jan 21, 2009

### Staff: Mentor

You sort of do. ln w5 = 5 ln w.

7. Jan 21, 2009

### prime-factor

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.