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∫tanx+cotx dx - Mathematics and Statistics

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प्रश्न

`int sqrt(tanx) + sqrt(cotx)  "d"x`

योग
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उत्तर

Let I = `int (sqrt(tanx) + sqrt(cotx))  "d"x`

= `int (sqrt(tanx) + 1/sqrt(tanx))  "d"x`

= `int (tanx + 1)/sqrt(tanx)  "d"x`

Put `sqrt(tanx)` = t

∴ tanx = t2

∴x = tan−1(t2)

∴ dx = `1/(1 + ("t"^2)^2) * 2"t" "dt"`

∴ dx = `(2"t")/(1 + "t"^4)  "dt"`

∴ I = `int ("t"^2 + 1)/"t"* (2"t")/(1 + "t"^4)  "dt"`

= `2 int ("t"^2 + 1)/("t"^4 + 1)  "dt"`

= `2 int (1 + 1/"t"^2)/("t"^2 + 1/"t"^2)  "dt"`

= `2 int (1 + 1/"t"^2)/(("t" - 1/"t")^2 + 2)`

Put `"t" - 1/"t"` = u

∴ `(1 + 1/"t"^2) "dt"` = du

∴ I = `2 int "du"/("u"^2 + 2)`

= `2 int  "du"/("u"^2 + (sqrt(2))^2`

= `2* 1/sqrt(2)tan^-1 ("u"/sqrt(2)) + "c"`

= `sqrt(2)tan^-1 (("t" - 1/"t")/sqrt(2)) + "c"`

= `sqrt(2)tan^-1 (("t"^2 - 1)/sqrt(2)) + "c"`

= `sqrt(2)tan^-1 ((tanx - 1)/sqrt(2tanx)) + "c"`

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अध्याय 2.3: Indefinite Integration - Long Answers III

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