InterviewSolution
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InterviewSolution
| 1. |
Evaluate `int(dx)/((sinx-sin2x)).` |
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Answer» `int(dx)/((sinx-sin2x))=int (dx)/((sin s-2 sin x cos x))` `=int (dx)/(sinx(1-2cos x ))=int(sinx)/(sin^(2)x(1-2 cos x))dx` `intj (sin x)/((1-cos^(2) x)(1-2cos x))dx` `=-int(dt)/((1-t^(2))(1-2t)),`where cos x=t `= int (dt)/((t-1)(t+1)(1-2t)).` `Let (1)/((t-1)(t+1)(1-2t))=(A)/((t-1))+(B)/((t+1))+( C)/((1-2t)).` `then , 1-=A(t+1)(1-2t)+B(t-1)(1-2t)+C(t-1)(t+1).` Putting `t=1` in (i) , we get `A=(-1)/(2)` Putting `t=-1` in (ii), we get `B=(-1)/(6).` Putting `t=(1)/(2)` in (ii), we get `C=(-4)/(3).` `therefore I=-(1)/(2)int(dt )/((t-1))-(1)/(6). int (dt) /((t+1))-(4)/(3).int (dt)/((1-2t))` `=-(1)/(2)log|t-1|-(1)/(6)log |t+1|+(2)/(3).int(-2dt)/((1-2t))` `=-(1)/(2) log |t-1|-(1)/(6)log |t+1|+(2)/(3)log|1-2t+C` `=-(1)/(2) log |cos x-1|-(1)/(6)log |cos x+1|+(2)/(3)log |1-2 cos x|+c.` |
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