1.

Evaluate:`int(tantheta+tan^3theta)/(1+tan^3theta)dtheta`

Answer» WE have
`((tan theta+tan^(3) theta))/((1+tan^(3) theta))=(tan theta(1+tan^(2) theta))/((1+tan^(3)theta))=(tan theta sec^(2)theta)/((1+tan^(3) theta)).`
`therefore I=int((tan theta+ tan ^(3) theta))/((1+tan^(3) theta))`
`= int (tan theta sec^(2) theta)/((1+tan^(3) theta))d theta`
`=int (t)/((1+t^(3)))dt=int (t)/((1+t)(1-t+t^(2))dt,`where `tan theta =t.`
`Let (t)/((1+t)(1-t+t^(2)))=(A)/((1+t))+((Bt+C))/((1-t+t^(2))).`then
`t-=A(1-t+t^(2))+(Bt+C)(1+t).`
Putting `t=-1` on both sides of (i) , we get `A=(-1)/(3).`
Comparing the coefficients of `t^(2)` on both sides of (i) , we get
`A+B=0impliesB=-A=(1)/(3).`
Comparing the constant terms on both sides of (i) , we get
`A+C=0impliesC=-A=(1)/(3).`
`therefore (t)/((1+t)(1-t+t^(2)))=(-1)/(3(1+t))+(((1)/(3)t+(1)/(3)))/((1-t+t^(2))).`
`Now,I=int(t)/((1+t)(1-t+t^(2)))dt`
`=-(1)/(3)int(dt )/((1+t))+(1)/(6)int(2t)/((t^(2)-t+1))dt+(1)/(3)int(dt)/((t^(2)-t+1))`
`=-(1)/(3)int(dt )/((1+t))+(1)/(6)int((2t-1)+1)/((t^(2)-t+1))dt+(1)/(3)int(dt)/((t^(2)-t+1))`
`=-(1)/(3)log|1+t|+(1)/(6)log|t^(2)+1|+(1)/(2)int(dt)/((t^(2)-t+(1)/(4))+(3)/(4))`
`=-(1)/(3)log|1+t|+(1)/(6)log|t^(2)+1|+(1)/(2)int (dt)/((t-1//2)^(2)+(sqrt(3)//2)^(2))`
`=-(1)/(3)log|1+t|+(1)/(6)log|t^(2)+t+1|+(1)/(2).(2)/(sqrt(3))tan^(-1)((t-(1)/(2)))/((sqrt(3)//2))+C`
`=-(1)/(3)log |1+t|+(1)/(6)log|t^(2)-t+1|+(1)/(sqrt(3))tan^(-1)((2t-1)/(sqrt(3)))+C`
`=-(1)/(3) log |1+ tan theta|+(1)/(6)log |tan^(2) theta - tan theta +1|+(1)/(sqrt(3))tan^(-1)((2 tan theta -1)/(sqrt(3)))+C.`


Discussion

No Comment Found

Related InterviewSolutions