#int cos^6x dx#
We can rewrite this function as;
#int (cos^2x)^3dx#
Now we can rewrite #cos^2 x# using the half angle formula.
#int(1/2(1+cos(2x)))^3 dx#
If we expand the expression we can get rid of the exponent.
#1/8 int (1 + 3cos(2x) + 3cos^2 (2x) + cos^3 (2x) )dx #
We can split up this expression using the sum rule.
#1/8 int dx + 3/8 int cos (2x) dx + 3/8 int cos^2 (2x) dx + 1/8 int cos^3 (2x) dx#
The first integral is pretty straight forward. For the second, use the substitution #u=2x, du = 2dx#.
#1/8x + 3/16 sin (2x) + 3/8 int cos^2 (2x) dx + 1/8 int cos^3 (2x) dx #
The remaining integrals are a little more involved.
We can use the half angle formula again to simplify the 3rd integral.
#int cos^2 (2x) dx = int 1/2(1+ cos(4x)) dx#
#= 1/2 int dx + 1/2 int cos(4x) dx#
Use substitution again to solve the second integral.
#1/2 x + 1/8 sin (4x) #
The last integral should be split up first.
# int cos^3 (2x) dx = int cos^2 (2x) cos (2x) dx #
Now we can use the Pythagorean theorem to replace #cos^2(2x)#.
#int (1-sin^2 (2x)) cos (2x) dx #
#int cos (2x) dx - int sin^2 (2x) cos (2x) dx #
#1/2 sin (2x) - int sin^2 (2x) cos (2x) dx #
To solve the remaining integral, use the substitution #u=sin (2x), du = 2 cos (2x) dx #.
#1/2 sin (2x) - 1/6 sin^3 (2x) #
Plug the solved integrals into the original expression to get;
#1/8x + 3/16 sin 2x + 3/8(1/2 x + 1/8 sin 4x ) + 1/8(1/2 sin 2x - 1/6 sin^3 2x)#
Simplify by combining like terms.
#5/16 x + 1/4 sin 2x + 3/16 sin 4x -1/48 sin^3 2x#
