Question

How do you find the antiderivative of `cos^4 x dx`?

Answer 1

to find this integration or anti-derivative we use Reduction formula

Explanation:

Reduction formula
`intcos^n xdx=(cos^(n-1) xsinx)/n +(n-1)/n intcos^(n-2) x dx`

using n=4 we get

`intcos^4 xdx=(cos^(4-1) xsinx)/4 +(4-1)/4 intcos^(4-2) x dx`
`intcos^4 xdx=(cos^(3) xsinx)/4 +(3)/4 intcos^(2) x dx`
`intcos^4 xdx=(cos^(3) xsinx)/4 +(3)/4 int(1/2 cos2x+1/2)dx)`
`intcos^4 xdx=(cos^(3) xsinx)/4 +(3)/8 int(cos2x)dx+3/8 int1dx)`
`intcos^4 xdx=(cos^(3) xsinx)/4 +(3)/8 sinx cosx+3/8 x`
`intcos^4 xdx=1/32 (12x+8sin(2x)+sin4x)+C`

that is the right answer.

Answer 2

`1/32{12x+8sin2x+sin4x}+C.`

Explanation:

Recall that, `cos2x=2cos^2x-1," so that, "cos^2x=(1+cos2x)/2.`

`:. cos^4x=((1+cos2x)/2)^2,`

`=1/4{1+2cos2x+cos^2(2x)},`

`=1/4{1+2cos2x+(1+cos4x)/2},`

`=1/8(3+4cos2x+cos4x).`

`rArr intcos^4xdx=1/8int(3+4cos2x+cos4x)dx,`

`=1/8{3x+4*sin(2x)/2+sin(4x)/4},`

`=1/32{12x+8sin2x+sin4x}+C.`