Evaluate: (i) `int(sin2x)/((a+bcos2x)^2)\ dx` (ii) `int(logx^2)/x\ dx`

Explanation

(i) `I=int(sin2x)/((a+bcos2x)^2)\ dx`
taking `a+bcos 2x =t`
`=>-2b sin 2xdx =dt`
`=>sin 2x dx =-1/(2b) dt`
so,` -1/(2b) int 1/t^2 dt`
`=-1/(2b) (-1/t)+c`
`=1/(2b) xx 1/t +c`
`=1/(2b) (1/(a+bcos2x))+c`

(ii) `I=int(logx^2)/x\ dx`
putting `t=log x^2`
`=>1/x^2(2x)dx=dt`
`=>2/x dx=dt`
`=>1/x dx=(dt)/2`
so,`I=1/2 int tdt`
`=1/2 xx t^2/2 +c`
`=t^2/4+c`
`=1/4 (log x^2)^2+c`