InterviewSolution
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InterviewSolution
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`int(dx)/((sinx - sin 2x))` का मान ज्ञात कीजिए । |
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Answer» `int(dx)/((sinx - sin 2x))= int(dx)/((sinx-2sin xcosx))` `=int(dx)/(sinx(1-2cosx))=int(sinx)/(sin^(2)x(1-2 cosx))dx` `=int(sinx)/((1-cos^(2)x)(1-2 cosx))dx` यदि `cos x = t ` तथा `-sin x dx=dt,` तब `=-int(dt)/((1-t^(2))(1-2t))=int(dt)/((t-1)(t+1)(1-2t))" ...(1)"` अब, आंशिक भिन्नो में व्यक्त करने पर `(1)/((t-1)(t+1)(1-2t))=(A)/((t-1))+(B)/((t+1))+(C)/((1-2t))" ...(2)"` `1=A(t+1)(1-2t)+B(t-1)(1-2t)+C(t-1)(t+1)` यदि `t=1," तब "A=-(1)/(2)` यदि `t=-1," तब "B=-(1)/(6)` और यदि `t=(1)/(2)," तब "C=-(4)/(3)` अब, समीकरण (1 ) व (2 ) से, `int(dt)/((t-1)(t+1)(1-2t))=-(1)/(2)int(dt)/((t-1))-(1)/(6)int(dt)/((t+1))` `=-(1)/(2)log(t-1)-(1)/(6)log(t+1)+(2)/(3)log(1-2t)+c` `=-(1)/(2)log(cosx-1)-(1)/(6)log(cosx+1)+(2)/(3)log(1- 2 cos x)+c` |
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