As bonus: lets calculate intcos^nxdx∫cosnxdx and then we will have an special case for n=4n=4
I_n=intcos^nxdxIn=∫cosnxdx lets do it by parts
u=cos^(n-1)xu=cosn−1x ; dv=cosxdxdv=cosxdx
du=(n-1)cos^(n-2)x(-sinx)dxdu=(n−1)cosn−2x(−sinx)dx and v=sinxv=sinx
I_n=sinxcos^(n-1)x+int(n-1)sin^2xcos^(n-2)x=In=sinxcosn−1x+∫(n−1)sin2xcosn−2x=
=sinxcos^(n-1)x+(n-1)int(1-cos^2x)cos^(n-2)xdx==sinxcosn−1x+(n−1)∫(1−cos2x)cosn−2xdx=
=sinxcos^(n-1)x+(n-1)intcos^(n-2)xdx-intcos^nxdx=sinxcosn−1x+(n−1)∫cosn−2xdx−∫cosnxdx
Then we have
I_n=sinxcos^(n-1)x+(n-1)I_(n-2)-I_nIn=sinxcosn−1x+(n−1)In−2−In
2I_n=sinxcos^(n-1)x+(n-1)I_(n-2)2In=sinxcosn−1x+(n−1)In−2
And we get the reduction formula
I_n=(sinxcos^(n-1)x)/2+(n-1)/2I_(n-2)In=sinxcosn−1x2+n−12In−2
n=0ddotsn=0⋱; I_0=intcos^0xdx=intdx=xI0=∫cos0xdx=∫dx=x
n=1ddotsn=1⋱ I_1=intcosxdx=sinxI1=∫cosxdx=sinx
n=2 ddotsn=2⋱; I_2=intcos^2xdx=(sinxcosx)/2+1/2I_0=(sinxcosx)/2+1/2xI2=∫cos2xdx=sinxcosx2+12I0=sinxcosx2+12x
n=3ddotsn=3⋱;I_3=(sinxcos^2x)/2+cancel2/cancel2sinx
n=4ddots;I_4=(sinxcos^3x)/2+3/2I_2=
=(sinxcos^3x)/2+3/2((sinxcosx)/2+1/2x)=
=(sinxcos^3x)/2+(3sinxcosx)/4+3/4x
And so on....
You can use other trigonometric identities to resume the final expresion...
