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प्रश्न
Evaluate the following:
`int sqrt(tanx) "d"x` (Hint: Put tanx = t2)
उत्तर
Let I = `int sqrt(tanx) "d"x`
Put tan x = t2
⇒ sec2x dx = 2t dt
∴ I = `int "t" * (2"t")/(sec^2x) "dt"`
= `2 int "t"^2/(1 + "t"^4) "dt"`
= `int (("t"^2 + 1) + ("t"^2 - 1))/((1 + "t"^4)) "dt"`
= `int ("t"^2 + 1)/(1 + "t"^4) "dt" + int ("t"^2 - 1)/(1 + "t"^4) "dt"`
= `int (1 + 1/"t"^2)/("t"^2 + 1/"t"^2) "dt" + int (1 - 1/"t"^2)/("t"^2 + 1/"t"^2) "dt"`
= `int (1 + 1/"t"^2)/(("t" - 1/"t")^2 + 2)"dt" + int (1 - 1/"t"^2)/(("t" + 1/"t")^2 - 2)"dt"`
Put u = `"t" - 1/"t"`
⇒ du = `(1 + 1/"t"^2)"dt"` in first integral
And put v = `"t" + 1/"t"`
⇒ dv = `(1 - 1/"t"^2)"dt"` in second integral
∴ I = `int "du"/("u"^2 + (sqrt(2)^2)) + int "dv"/("v"^2 - (sqrt(2)^2))`
= `1/sqrt(2) tan^-1 "u"/sqrt(2) + 1/(2sqrt(2)) log|("v" - sqrt(2))/("v" + sqrt(2))| + "C"`
= `1/sqrt(2) tan^-1 ("t" - 1/"t")/sqrt(2) + 1/(2sqrt(2)) log |("t" + 1/"t" - sqrt(2))/("t" + 1/"t" + sqrt(2))| + "C"`
= `1/sqrt(2) tan^-1 ("t"^2 - 1)/(sqrt(2)"t") + 1/(2sqrt(2)) log |("t"^2 + 1 - sqrt(2)"t")/("t"^2 + 1 + sqrt(2)"t")| + "C"`
= `1/sqrt(2) tan^-1 ((tanx - 1)/sqrt(2tan x)) + 1/(2sqrt(2)) log |(tan x - sqrt(2 tanx) + 1)/(tan x + sqrt(2 tan x) + 1)| + "C"`
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