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∫[cosec(logx)][1-cot(logx)] dx

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प्रश्न

`int ["cosec"(logx)][1 - cot(logx)]  "d"x`

योग
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उत्तर

Let I = `int ["cosec"(logx)][1 - cot(logx)]  "d"x`

Put logex = t

∴ x = et

∴ dx = `"e"^"t"*"dt"`

∴ I = `int "cosec"  "t"(1 - cot "t")  "e"^"t"  "dt"`

= `int "e"^"t" ("cosec"  "t" - "cosec"  "t"*cot "t")  "dt"`

Put f(t) = cosec t

∴ f'(t) = −cosec t.cot t

∴ I = `int"e"^"t"  ["f"("t") + "f'"("t")]  "dt"`

= et ⋅ f(t) + c = et cosec t + c

∴ I = `x  "cosec"  (logx) + "c"`

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अध्याय 2.3: Indefinite Integration - Short Answers I

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