Write the integrand as:
cos^6x = cos^5x * cosx cos6x=cos5x⋅cosx
So:
int cos^6x dx = int cos^5x * cosx dx∫cos6xdx=∫cos5x⋅cosxdx
Integrate by parts:
int cos^6x dx = int cos^5x d/dx (sinx) dx∫cos6xdx=∫cos5xddx(sinx)dx
int cos^6x dx = cos^5x sinx - int sinx d/dx (cos^5x) dx∫cos6xdx=cos5xsinx−∫sinxddx(cos5x)dx
int cos^6x dx = cos^5x sinx + 5 int sin^2x cos^4x dx∫cos6xdx=cos5xsinx+5∫sin2xcos4xdx
Use now the identity: sin^2x = 1-cos^2xsin2x=1−cos2x
int cos^6x dx = cos^5x sinx + 5 int ( 1-cos^2x) cos^4x dx∫cos6xdx=cos5xsinx+5∫(1−cos2x)cos4xdx
using the linearity of the integral:
int cos^6x dx = cos^5x sinx + 5 int cos^4x dx - 5intcos^6x dx∫cos6xdx=cos5xsinx+5∫cos4xdx−5∫cos6xdx
The integral now appears on both sides of the equation:
6 int cos^6x dx = cos^5x sinx + 5 int cos^4x dx 6∫cos6xdx=cos5xsinx+5∫cos4xdx
int cos^6x dx = ( cos^5x sinx )/6 + 5/6 int cos^4x dx ∫cos6xdx=cos5xsinx6+56∫cos4xdx
Using the same method we can find that:
int cos^4x dx = ( cos^3x sinx )/4 + 3/4 int cos^2x dx ∫cos4xdx=cos3xsinx4+34∫cos2xdx
int cos^2x dx = ( cosx sinx )/2 + 1/2 int dx = (cosxsinx+x)/2+C∫cos2xdx=cosxsinx2+12∫dx=cosxsinx+x2+C
Putting together the partial results:
int cos^6x dx = ( cos^5x sinx )/6 + 5/24 ( cos^3x sinx ) +15/48(cosxsinx+x)+C∫cos6xdx=cos5xsinx6+524(cos3xsinx)+1548(cosxsinx+x)+C
and simplifying:
int cos^6x dx = ( 8cos^5x sinx +10 cos^3x sinx +15cosxsinx+15x)/48+C∫cos6xdx=8cos5xsinx+10cos3xsinx+15cosxsinx+15x48+C