InterviewSolution
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InterviewSolution
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`int_(0)^(1)(x tan^(-1)x)/((1+x^(2))^(3//2))dx` का मान ज्ञात कीजिए । |
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Answer» `int_(0)^(1)(xtan^(-1)x)/((1+x^(2))^(3//2))dx` यदि `tan theta= x rArr sec^(2)theta d theta = dx` तथा सीमायें हैं। `theta=0` पर `x=0` व `theta=pi//4` पर `x = 1,` तब `int_(0)^(1)(x tan^(-1))/((1+x^(2))^(3//2))dx=int_(0)^(pi//4)(theta tan theta)/(sec^(3)theta).sec^(2) theta d theta` `=int_(0)^(pi//4)theta sin theta d theta` खण्डशः समाकलन करने पर , `=[-theta cos theta]_(0)^(pi//4)-int_(0)^(pi//4)(-cos theta) d theta` `=[-theta-cos theta]_(0)^(pi//4)+[sinx theta]_(0)^(pi//4)` `=-(pi)/(4)cos.(pi)/(4)+sin.(pi)/(4)` `=((-pi)/(4sqrt2)+(1)/(sqrt2))=((4-pi))/(4sqrt2)=(sqrt2(4-pi))/(8)` |
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