InterviewSolution
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InterviewSolution
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`int(dx)/(sinx+sin2x)` का मान ज्ञात कीजिए । |
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Answer» यदि `" " I=int(dx)/(sinx+sin2x)` `=int(dx)/(sinx+2 sinx cosx)` `=int(dx)/(sinx(1+2 cosx))` माना `cos x = t rArr - sin x dx = dt` तब `" "I=int(-(dt)/(sinx))/(sinx(1+2t))=-int(dt)/(sin^(2)x(1+2t))` `=-int(dt)/((1-t^(2))(1+2t))" ...(1)"` समीकरण `(1)/((1-t^(2))(1+2t))` को आंशिक भिन्नों में व्यक्त करने पर , `(1)/((1-t^(2))(1+2t))=(A)/(1-t)+(B)/(1+t)+(C)/(1+2t)` `rArr 1 = A(1+t)(1+2t)+B(1-t)(1+2t)+c(1-t)(1+t)" ...(2)"` समीकरण (2 ) में `t=1, -1` व `-(1)/(2)` रखने पर `A=(1)/(6), B=-(1)/(2),C=(4)/(3)` तब समीकरण (1 ) व (2 ) से `I=-int[(1)/(6).(1)/(1-t)-(1)/(2).(1)/(1+t)+(4)/(3).(1)/(1+2t)]dt` `=-[-(1)/(6)log(1-t)-(1)/(2)log|1+t|+(4)/(3).(1)/(2)log|1+2t|+c]` `=(1)/(6)log|1-cos x|+(1)/(2)log|1+cosx|-(2)/(3)log|1+2cos x|+c` |
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