Question

What is `int sin^(2)t cos^(4)t dt`?

Answer 1

`int1/8*(2sintcost)^2*(2cos^2t)dt`
`=int1/8*sin^2 2t*(1+cos2t)dt`
`=int1/16*2sin^2 2t*(1+cos2t)dt`
`=int1/16*(1-cos4t)*(1+cos2t)dt`

`=1/16int(1-cos4t)*(1+cos2t)dt`

`=1/16[intdt+intcos2tdt-intcos4tdt-1/2int(2cos4tcos2tdt)]`
`=1/16[intdt+intcos2tdt-intcos4tdt-1/2int(cos6t+cos2t)dt]`

`=1/16[t+1/2sin2t-1/4sin4t-1/12sin6t-1/4sin2t+c]`
`=1/16[t+1/4sin2t-1/4sin4t-1/12sin6t+c]`