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प्रश्न
`int sin(logx) "d"x`
उत्तर
Let I = `int sin(log x) "d"x`
Put log x = t
∴ x = et
∴ dx = et dt
∴ I = `int sin "t" * "e"^"t" "dt"`
= `sin "t" int "e"^"t" "dt" - int ["d"/"dt" (sin "t") int "e"^"t" "dt"]"dt"`
= `sin "t"* "e"^"t" - int cos "t" * "e"^"t" "dt"`
= `"e"^"t" sin "t" - [cos "t" int "e"^"t" "dt" - int ("d"/"dt"(cos "t") int "e"^"t" "dt")"dt"]`
= `"e"^"t" sin "t" - ["e"^"t" cos "t" - int(- sin "t")"e"^"t" "dt"]`
= `"e"^"t" sin "t" - "e"^"t"cos "t" - int sin "t" * "e"^"t" "dt"`
∴ I = `"e"^"t"(sin "t"- cos "t") - "I" + "c"_1`
∴ 2I = `"e"^"t"(sin "t" - cos "t") + "c"_1`
∴ I = `"e"^"t"/2 (sin "t" - cos "t") + "c"_1/2`
∴ I = `x/2 [sin (log x) - cos(log x)] + "c"`,
where c = `"c"_1/2`
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