Name:
#I_4 = int cos^4x dx#
and integrate it by parts:
#I_4 = int cos^3x cosx dx = int cos^3x d(sinx) = sinxcos^3x + 3 int cos^2xsin^2xdx#
Using the identity: # sin^2x = 1-cos^2x# we have:
#I_4 = sinxcos^3x + 3 int cos^2x (1- cos^2x) dx#
#I_4 = sinxcos^3x + 3 int cos^2xdx -3 int cos^4x dx#
#I_4 = sinxcos^3x + 3 int cos^2xdx -3 I_4#
Solving for #I_4#:
#I_4 = (sinxcos^3x)/4 + 3/4 int cos^2xdx#
Use now the identity:
#cos^2x = (1+cos 2x)/2#
so:
#int cos^2xdx = int (1+cos 2x)/2dx = 1/2 int dx + 1/2int cos2x dx = 1/2x+ 1/4 sin 2x +C = 1/2(x+sinxcosx)+C#
and putting it together:
#int cos^4x dx = (3x+ 2sinxcos^3x+3sinxcosx)/8+C#
Considering that:
#sin 2x = 2sinxcosx#
#sin 4x = 2sin2xcos2x = 4sinxcosx(2cos^2x-1) = 8sinxcos^3x -4sinxcosx#
we can also write the solution as:
#int cos^4x dx = (12x+ 8sinxcos^3x-4sinxcosx+16sinxcosx)/32+C#
#int cos^4x dx = (12x+ 8sin 2x +sin 4x)/32+C#
