Expert: Paul Klarreich - 9/28/2008

Question
Trying to integrate cos^4 x dx  I,ve been told a Hint
cos^4x  = cos^2 x X cos^2 x  and cos^2 x = (1 + cos 2x )/2
I,ve tryed all substitutions I can to no avail
I know answer is 3/8 x + 1/4 sin 2x + 1/32 sin 4x  + c
I always end up with a Sin I don't want

Answer
Questioner:   carole
Category:  Calculus
Private:  No
 
Subject:  Integration
Question:  Trying to integrate cos^4 x dx  I've been given a Hint
cos^4x  = cos^2 x X cos^2 x  and cos^2 x = (1 + cos 2x )/2
I've tried all substitutions I can but to no avail
I know answer is 3/8 x + 1/4 sin 2x + 1/32 sin 4x  + c
I always end up with a Sin I don't want
..............................................
Hi, Carole,

Suggestion for the future: Check the archives. Click BROWSE PAST ANSWERS.  You'll find this one -- you're not the first to run into it.

This is called the 'half-angle trick':
             1 + cos t
cos^2 (t/2) = ----------
                2

Since you have cos^4 x, you must use it twice.

cos^4(x) = (cos^2(x))^2

 (1 + cos 2x)^2
= --------------
       4

 1 + 2 cos(2x) + cos^2(2x)
= ---------------------------
              4

Now use it again:

 1 + 2 cos(2x) + (1 + cos(4x)) /2
= ---------------------------
              4

Clear fractions:


 2 + 4 cos(2x) + 1 + cos(4x)
= ---------------------------
              8

 3 + 4 cos(2x) + cos(4x)
= ------------------------
            8

I think you can handle it from here.