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`int(sinx)/((1+cosx)(2+3cosx))dx` का मान ज्ञात कीजिए ।

Answer» माना `I=int(sinx)/((1+cosx)(2+3cosx))dx`
`=int(-dt)/((1+t)(2+3t))`
`" माना "cosx=t`
`" "-sinx=(dt)/(dx)`
`" "rArr sinx dx =-dt`
माना `(-1)/((1+t)(2+3t))=(A)/(1+t)+(B)/(2+3t)=(A(2+3t)+B(1+t))/((1+t)(2+3t))`
`rArr" "A(2+t)+B(1+t)=-1`
`t=-1` रखने पर
`A(2-3)+0=-1`
`rArr" "A=1`
`t=-(2)/(3)` रखने पर
`0+B(1-(2)/(3))=-1`
`rArr" "B=-3`
`therefore" "I=int((1)/(1+t)-(3)/(2+3t))dt`
`=log|1+t|-log|2+3t|+c`
`=log|(1+t)/(2+3t)|+c`
`=log|(1+cosx)/(2+3cosx)|+c`


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