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प्रश्न
`int sqrt(tanx) + sqrt(cotx) "d"x`
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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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