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InterviewSolution
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`int (d theta)/(sin theta +sin 2theta)` |
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Answer» Let `I= int (d theta)/(sin theta + sin 2 theta)` `= int (d theta)/(sint heta +2 sin theta* cos theta)` `=int (d theta)/(sin theta (1+2cos theta))` `= int (sin theta * d theta)/(sin^(2) theta*(1+2cos theta))` `= int (sin theta*d theta)/((1-cos^(2) theta) *(1+2 cos theta))` `= int(sin theta* d theta)/((1+cos theta)(1-cos theta)(1+2 cos theta))` Let `cos thetat= t :. sin theta * d theta = dt` `:. I= int (-dt)/((1+t)(1-t)(1+2t))` Consider `(-1)/((1+t)(1-t)(1+2t))=(A)/(1+t)+(B)/(1-t)+(C)/(1+2t)` `:. -1A(1-t)(1+2t)+B(1+t)(1+2t)+C(1+t)(1-t)` Put t=1 `:. -1=B(2xx3)` `:. B=(-1)/(6)` Put `t=-1` `:. -1=a(2)(-1)` `:. A=(1)/(2)` Put `t=-(1)/(2)` `:. -1 =C (-1(1)/(2))(1+(1)/(2))` `:. -1=C((1)/(2))((3)/(2))=C ((3)/(4))` `:. C=(-4)/(3)` `:. (-1)/((1+t)(1-t)(1+2r))=((1)/(2))/(1+t)+(-(1)/(6))/(1-t)+(-(4)/(3))/(1+3t)` `:. I= int (-dt)/((1+t)(1-t)(1+2t))` `= int [(1)/(2(1+t))-(1)/(6(1-t))-(4)/(3(1+2t))]*dt` `=(1)/(2)int (1)/(1+t)*dt -(1)/(6) int (1)/(1-t)dt-(4)/(3)int (1)/(1+2t)dt` `=(1)/(2) log |(1+t)|+(1)/6)log |(1-t)|-(4)/(3xx2)log|(1+2t)|+c` `I=(1)/(2)log|1+cos theta|+(1)/(6)log|` `1-cos theta| -(2)/(3)log |1+2cos theta|+c` |
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