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प्रश्न
`int 1/(2 + cosx - sinx) "d"x`
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उत्तर
Let I = `int 1/(2 + cosx - sinx) "d"x`
Put `tan (x/2)` = t
∴ x = 2 tan−1t
∴ dx = `(2"dt")/(1 + "t"^2)` and sin x = `(2"t")/(1 + "t"^2)`,cos x = `(1 - "t"^2)/(1 +"t"^2)`
∴ I = `int 1/(2 + ((1 - "t"^2)/(1 + "t"^2)) - (2"t")/(1 + "t"^2)) xx (2"dt")/(1 + "t"^2)`
= `int 2/(2+ 2"t"^2 + 1 - "t"^ - 2"t") "dt"`
= `2 int 1/("t"^2 - 2"t" + 3) "dt"`
`(1/2 "coefficient of t")^2 = (1/2 xx -2)^2`
= `(-1)^2`
= 1
∴ I = `2 int 1/("t"^2 - 2"t" + 1 - 1 + 3) "dt"`
= `2int 1/(("t" - 1)^2 + 2) "dt"`
= `2int 1/(("t" - 1)^2 + (sqrt(2))^2) "dt"`
= `2*1/sqrt(2) tan^(-1) (("t" - 1)/sqrt(2))+ "c"`
∴ I = `sqrt(2) tan^(-1) [(tan(x/2) - 1)/sqrt(2)] + "c"`
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