`(cosx)/((1-sinx)(2-sinx))`

Explanation

माना `I=int(cosx)/((1-sinx)(2-sinx))dx`
माना `sinx=t" "rArr" "cos x=(dt)/(dx)" "rArr" "cos dx =dt`
`I=int(1)/((1-t)(2-t))=int(1)/((1-t)(2-t))dt`
माना `" "(1)/((1-t)(2-t))=(A)/(1-t)+(B)/(2-t)" …(1)"`
`rArr" "(1)/((1-t)(2-t))=(A(2-t)+B(1-t))/((1-t)(2-t))`
`rArr" "1=A(2-t)+B(1-t)`
`{:(t=1,"तो ",1=A(2-1)+0,rArr.,A=1),(t=2,"तो ",1=0+B(1-2),rArr.,B=-1):}`
`therefore" "I=int((1)/(1-t)-(1)/(2-t))dt" "` [ समीकरण (1 ) से ]
`" "=int(1)/((1-t))dt-int(1)/((2-1))dt`
`" "=(log|1-t|)/((-1))-(log|2-t|)/((-1))+C`
`" "=log|(2-t)/(1-t)|+C=log|(2-sinx)/(1-sinx)|+C`