What is the integral of cosine^6 (x) dx ?

Write the integrand as:

#cos^6x = cos^5x * cosx #

So:

#int cos^6x dx = int cos^5x * cosx dx#

Integrate by parts:

#int cos^6x dx = int cos^5x d/dx (sinx) dx#

#int cos^6x dx = cos^5x sinx - int sinx d/dx (cos^5x) dx#

#int cos^6x dx = cos^5x sinx + 5 int sin^2x cos^4x dx#

Use now the identity: #sin^2x = 1-cos^2x#

#int cos^6x dx = cos^5x sinx + 5 int ( 1-cos^2x) cos^4x dx#

using the linearity of the integral:

#int cos^6x dx = cos^5x sinx + 5 int cos^4x dx - 5intcos^6x dx#

The integral now appears on both sides of the equation:

#6 int cos^6x dx = cos^5x sinx + 5 int cos^4x dx #

# int cos^6x dx = ( cos^5x sinx )/6 + 5/6 int cos^4x dx #

Using the same method we can find that:

# int cos^4x dx = ( cos^3x sinx )/4 + 3/4 int cos^2x dx #

# int cos^2x dx = ( cosx sinx )/2 + 1/2 int dx = (cosxsinx+x)/2+C#

Putting together the partial results:

# int cos^6x dx = ( cos^5x sinx )/6 + 5/24 ( cos^3x sinx ) +15/48(cosxsinx+x)+C#

and simplifying:

# int cos^6x dx = ( 8cos^5x sinx +10 cos^3x sinx +15cosxsinx+15x)/48+C#

Assuming that you wanted to type #intcos^6(x)dx#, we have to linearized #cos^6(x)#

#I==int(1/32(cos(6x)+6cos(4x)+15cos(2x)+10))dx#

#=1/32intcos(6x)dx+3/16intcos(4x)dx+15/32intcos(2x)+5/16intdx#

#=1/192int6cos(6x)dx+3/64int4cos(4x)dx+15/64int2cos(2x)dx+#

For each integral, we will use variable,

let #X=6x#, #Y=4x#, #Z=2x#,
#dX=6dx#, #dY=4dx#, #dZ=2dx#

So :

#I=1/192intcos(X)dX+3/64intcos(Y)dY+15/64intcos(Z)dZ+5//16x#
#=1/192sin(X)+3/64sin(Y)+15/64sin(Z)+5/16x+C#, #C in RR#

#=1/192sin(6x)+3/64sin(4x)+15/64sin(2x)+5/16x+C#, #C in RR#

\0/ here's our answer !