Question

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

Answer 1

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

Explanation:

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`

Answer 2

`intcos^6(x)dx=1/192sin(6x)+3/64sin(4x)+15/64sin(2x)+5/16x+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 !