Evaluate: `int(cos2xsin4x dx)/(cos^4x(1+cos^2 2x)`

Answer

`-(1)/(2)log|cos 2x|-(1)/(2(1+cos 2x))+(1)/(4)log(1+cos^(2)2x)+C`

Explanation

`I=int(cos2x sin4x dx)/(cos^(4)x(1+cos^(2)2x))`
`=int(2cos^(2)2x sin2x dx)/(((1+cos2x)/(2))^(2)(1+cos^(2) 2x))`
`"Let " cos2x=t. " then "dt=-2sin2x dx. `
`:. I=-int(t^(2)dt)/(((1+t)/(2))^(2)(1+t^(2)))=-4int(t^(2)dt)/((1+t)^(2)(1+t^(2)))`
` "Now ",(t^(2))/((1+t)^(2)(1+t^(2)))=(A)/(1+t)+(B)/((1+t)^(2)) +(Ct+D)/(1+t^(2))`
`"or " t^(2)=A(1+t)(1+t^(2))+B(1+t^(2))+(Ct+D)(1+t)^(2) `
` "Put " t=-1. " Then "B=1//2`
`"Put "t=0. " Then "0=A+1//2+D" (1)" `
`"Put "t=1. " Then " 1=4A+1+4C+4D " or "A+C+D=0 " "(2)`
`"From equations(1)and(2), " C=1//2`
`"Comparing coefficients of "t^(3),A+C=0 " "(3)`
`"or "A=-1//2`
`"From equations(2)and(3)",D=0`
`"Hence, " I=int(-(1//2)/(1+t)+(1//2)/((1+t)^(2))+((1//2)t)/(1+t^(2)))dt`
`=-(1)/(2)log|t|-(1)/(2(1+t))+(1)/(4)log(1+t^(2))+C`
`=-(1)/(2)log|t|-(1)/(2(1+t))+(1)/(4)log(1+t^(2))+C," where "t=cos2x`