how would one integrate an even trigonometric function such as cos^4 (x) dx.I tried breaking cos down into two cos functions, but failed to find a solution. please help me

12/5/2005 3:12:59 PM

1/4*cos(x)^3*sin(x)+3/8*cos(x)*sin(x)+3/8*x

That is the answer basiclly its insaine to do it by hand so put it into maple or a ti-89. If someone else has a better answer by doing it my hand or something. I would like to know too.

12/5/2005 3:22:16 PM

break cos to two square functions, then use

Cos^2 (x) = (1 + cos (2x))/2 for both of them.

12/5/2005 3:23:25 PM

[cos(x)]^4 = [ cos^2(x) ]^2
= [ (1\2)(1+cos(2x) ) ]^2
= (1/4)[ 1 + 2cos(2x) + cos^2(2x) ]
= (1/4)[ 1 + 2cos(2x) + (1\2)(1+cos(4x) ]
= (3/8)+(1/2)cos(2x) + (1/8)cos(4x)

Just applied the double angle formula twice. Of course you still need to integrate but that should be simple in view of the algebra above.

Alternatively you could use the imaginary exponentials to arrive at the same algebra, or use the recurrence relation of integrals of cosine.

12/5/2005 3:24:21 PM

^^^ I welcome this kind of insanity into my life.

[Edited on December 5, 2005 at 3:26 PM. Reason : can't count]

12/5/2005 3:25:50 PM

cos2x = 1 + cos 2x/2

cos4x = ((1+ cos2x) (1+cos 2x))/4

? cos4x dx = ? ((1+ cos2x) (1+cos 2x))/4 dx

= ? (1 + cos22x + 2cos2x)/4 dx

= (1/4) ?dx + ? cos22x dx + (1/2)?cos2x dx

= (1/4) ?dx + ? ((1 + cos 2x)/2)dx + (1/2)?cos2x dx

and integrate (replace question mark with integral sign)




[Edited on December 5, 2005 at 3:32 PM. Reason : .]

12/5/2005 3:30:32 PM

Quote :

"I welcome this kind of insanity into my life."

God save you! haha

12/5/2005 3:35:59 PM