InterviewSolution
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InterviewSolution
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समाकलनों`I_(1)=int_(pi//6)^(pi//3)(dx)/(1+sqrt(tanx))` और `I_(2)=int_(pi//6)^(pi//3)(sqrt(sinx)dx)/(sqrt(sinx)+sqrt(cosx))` पर विचार कीजिए। `I_(1)-I_(2)` किसके तुल्य है ? |
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Answer» Correct Answer - A दिया है, `I_(1)=int_(pi//6)^(pi//3)(dx)/(1+sqrt(tanx))` तथा `I_(2)=int_(pi//6)^(pi//3)(sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx` . . . (i) `I_(1)` को निम्न प्रकार लिखा जा सकता है `I_(1)=int_(pi//6)^(pi//3)(dx)/(1+(sqrt(sinx))/(sqrt(cosx)))` `=int_(pi//6)^(pi//3)(sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx`. . . (ii) अब, `I_(1)-I_(2)=int_(pi//6)^(pi//3)(sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx` `-int_(pi//6)^(pi//3)(sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx` `=int_(pi//6)^(pi//3)((sqrt(cosx)-sqrt(sinx))/(sqrt(cosx)+sqrt(sinx)))dx` `=int_(pi//6)^(pi//3)` `(sqrtcos((pi)/(3)+(pi)/(6)-x)-sqrt(sin((pi)/(3)+(pi)/(6)-x)))/(sqrt(cos((pi)/(3)+(pi)/(6)-x))+sqrt(sin((pi)/(3)+(pi)/(6)-x)))dx` `[becauseint_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx]` `=int_(pi//6)^(pi//3)(sqrt(cos((pi)/(2)-x))-sqrt(sin((pi)/(2)-x)))/(sqrt(cos((pi)/(2)-x))+sqrt(sin((pi)/(2)x)))dx` `=int_(pi//6)^(pi//3)(sqrt(sinx)-sqrt(cosx))/(sqrt(sinx)+sqrt(cosx))dx` `=-{int_(pi//6)^(pi//3)(sqrt(cosx)-sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx}` `=-(I_(1)-I_(2)rArr2(I_(1)-I_(2))=0` `rArrI_(1)-I_(2)=0` |
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