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सिद्ध कीजिये - `int_(0)^(pi//2)(dx)/(1+tanx)dx=(pi)/(8)log2`

Answer» `I=int_(0)^(pi//2)(dx)/(1+tanx)=int_(0)^(pi//2)(cosx)/(cosx+sinx)dx" …(1)"`
`=int_(0)^(pi//2)(cos((pi)/(2)-x))/(cos((pi)/(2)-x)+sin((pi)/(2)-x))dx`
`=int_(0)^(pi//2)(sinx)/(sinx+cosx)dx" ...(2)"`
समीकरण (1 ) व (2 ) को जोड़ने पर
`rArr" "2I=int_(0)^(pi//2)(sinx+cosx)/(sinx+cosx)dx=int_(0)^(pi//2)1.dx`


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