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प्रश्न
Find:
`int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`
उत्तर
Let
` "I" = int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`
` "I" = int_(-pi/4)^0 (1+sin"x"/cos"x")/(1-sin"x"/cos"x")"dx"`
` "I" = int_(-pi/4)^0 (cos"x"+sin"x")/(cos"x"-sin"x")"dx"`
Let
cos x - sin x = t
- (sin x + cos x) dx = dt
For x `=(-pi)/4,"t" = sqrt2` and x = 0, t = 1
`"I" = -int_(sqrt2)^1 "dt"/"t"`
`"I" = int_1^sqrt2 "dt"/"t"`
` = ["lnt"]_1^sqrt2`
` = "ln"sqrt2`
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