Expert: Paul Klarreich - 10/8/2007

Question
Hey Paul or Mr Klarreich (whatever you prefer) thanks for helping me. So I have four difficult integration problems. Im supposed to integrate these:
sin(4x)^4 dx

(3x^3-7x^2)/(3x+2) dx

(1+cos(x))^(1/2) dx

dx/(x(log8(x))dx (the base of the log is 8)

thanks a lot for your help

Answer
Questioner:   Jeff
Category:  Calculus
Private:  No
 
Subject:  integration problems
Question:  Hey Paul or Mr Klarreich (whatever you prefer) thanks for helping me. So I have four difficult integration problems. Im supposed to integrate these:
sin(4x)^4 dx

(3x^3-7x^2)/(3x+2) dx

(1+cos(x))^(1/2) dx

dx/(x(log8(x))dx (the base of the log is 8)

thanks a lot for your help
..................................
Hi, Jeff,

This is a lot of questions, so I'll try to get you started on each and leave the details to you.
=======================================================
WARNING: USE COURIER FONT TO VIEW THIS
===================================================

............................
{
| sin(4x)^4 dx
}


For even powers of sine or cosine, use the half-angle trick:
          1 - cos 2u
sin^2(u) = ------------
              2

{
| sin(4x)^4 dx =
}

{
| (sin(4x)^2)^2 dx =
}


{ (1 - cos 8x)^2
| --------------- dx
}       4

{ 1 - 2 cos 8x + cos^2(8x)
| ------------------------- dx
}           4

Now you have three terms.  The first two are routine, but the third is another even power, and you use the half-angle trick again:

          1 + cos 2u
cos^2(u) = ------------
              2
..............................................
{  (3x^3-7x^2)
|  ----------- dx
}    (3x+2)


When you have a 'big' polynomial on top, and a 'smaller' one on the bottom, you use long division first:


3x + 2  )  3x^3 - 7x^2          (  x^2 + 3x - 2
          3x^3 + 2x^2
          -----------
                 9x^2
                 9x^2 + 6x
                ------------
                      - 6x
                      - 6x - 4
                    ------------
                           + 4, the remainder.

Now the fraction is:
                  4
x^2 + 3x - 2 +  -------
               3x + 2

There should be no problem now.  Integrate the polynomial and the fraction.  The fraction gives you a  ln(..).

..................................

{
| sqrt(1 + cos x) dx
}

How about rationalizing the numerator?  You only learned rationalizing a denominator in high school?   You're in college now -- you're allowed.

sqrt(1 + cos x)  sqrt(1 - cos x)
--------------- ----------------- =
      1         sqrt(1 - cos x)

sqrt(1 - cos^2(x))
-------------------
sqrt(1 - cos x)

sqrt(sin^2(x))
---------------
sqrt(1 - cos x)

   sin(x)
---------------
sqrt(1 - cos x)

Now do a basic  u-substitution:

u = cos x,  du = - sin x dx,

and you have:

{     - du
|  ------------
}  sqrt(1 - u)

Now we can try a 'rationalizing substitution.'

Let  y = sqrt(1 - u)
y^2 = 1 - u

2y dy = - du

{     2y dy
|  ------------
}      y

{
| 2 dy
}

2y = etc....

.............................

dx/(x(log8(x))dx

Did you mean:

{    dx
|  -------
}  x ln x

[We can worry about the log8(x) later.]

How about  u = ln x?  du = dx/x,   dx = x du

{   x du
|  -------
}   x u

{   du
|  ----
}    u

= ln u = ln(ln x)

Now what about log8(x)?  Since  log8(x) = ln(x)/ln(8), all we need is a factor of  ln(8) in there.