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∫12+ cosx-sinx dx

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प्रश्न

`int 1/(2 +  cosx - sinx)  "d"x`

योग
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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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अध्याय 2.3: Indefinite Integration - Short Answers II

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