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

2 Answers
Jul 29, 2018

int cos^6x dx = ( 8cos^5x sinx +10 cos^3x sinx +15cosxsinx+15x)/48+Ccos6xdx=8cos5xsinx+10cos3xsinx+15cosxsinx+15x48+C

Explanation:

Write the integrand as:

cos^6x = cos^5x * cosx cos6x=cos5xcosx

So:

int cos^6x dx = int cos^5x * cosx dxcos6xdx=cos5xcosxdx

Integrate by parts:

int cos^6x dx = int cos^5x d/dx (sinx) dxcos6xdx=cos5xddx(sinx)dx

int cos^6x dx = cos^5x sinx - int sinx d/dx (cos^5x) dxcos6xdx=cos5xsinxsinxddx(cos5x)dx

int cos^6x dx = cos^5x sinx + 5 int sin^2x cos^4x dxcos6xdx=cos5xsinx+5sin2xcos4xdx

Use now the identity: sin^2x = 1-cos^2xsin2x=1cos2x

int cos^6x dx = cos^5x sinx + 5 int ( 1-cos^2x) cos^4x dxcos6xdx=cos5xsinx+5(1cos2x)cos4xdx

using the linearity of the integral:

int cos^6x dx = cos^5x sinx + 5 int cos^4x dx - 5intcos^6x dxcos6xdx=cos5xsinx+5cos4xdx5cos6xdx

The integral now appears on both sides of the equation:

6 int cos^6x dx = cos^5x sinx + 5 int cos^4x dx 6cos6xdx=cos5xsinx+5cos4xdx

int cos^6x dx = ( cos^5x sinx )/6 + 5/6 int cos^4x dx cos6xdx=cos5xsinx6+56cos4xdx

Using the same method we can find that:

int cos^4x dx = ( cos^3x sinx )/4 + 3/4 int cos^2x dx cos4xdx=cos3xsinx4+34cos2xdx

int cos^2x dx = ( cosx sinx )/2 + 1/2 int dx = (cosxsinx+x)/2+Ccos2xdx=cosxsinx2+12dx=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)+Ccos6xdx=cos5xsinx6+524(cos3xsinx)+1548(cosxsinx+x)+C

and simplifying:

int cos^6x dx = ( 8cos^5x sinx +10 cos^3x sinx +15cosxsinx+15x)/48+Ccos6xdx=8cos5xsinx+10cos3xsinx+15cosxsinx+15x48+C

Jul 29, 2018

intcos^6(x)dx=1/192sin(6x)+3/64sin(4x)+15/64sin(2x)+5/16x+Ccos6(x)dx=1192sin(6x)+364sin(4x)+1564sin(2x)+516x+C, C in RR

Explanation:

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 !