InterviewSolution
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सिद्ध कीजिएः की `int_(0)^(1) tan^(-1)xdx=pi/4-1/2log2.` |
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Answer» माना `I=underset(0)overset(1)int tan^(-1)xdx " "x=tan theta ` रखने पर `dx=sec^(2) theta d theta` जब x=0 तब `theta=0` तथा जब x=1 तब `theta=pi/4` `therefore I=underset(0)overset(pi/4)int tan^(-1)(tan theta) sec^(2) theta d theta ` `=underset(0) overset(pi/4)int theta sec^(2) d theta` `=[theta int theta sec^(2) d theta]_(0)^(pi)-underset(0)overset(pi/4)int {(d)/(d theta) intsec^(2) theta d theta}d theta` `=[theta tan theta]_(0)^(pi//4)-underset(0)overset(pi//4)int tan theta d theta` `=[pi/4tan""pi/4-0]+[log cos theta]_(0)^(pi//4)` `=pi/4. 1 log cos""pi/4-log cos theta` `=pi/4-log ""1/sqrt2-log 0` `[therefore log 1=0" तथा "pi/4log ""1/A=-logA]` `therefore I=pi/4-1/2log2 ` |
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