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∫sin(logx) dx

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Question

`int sin(logx)  "d"x`

Sum
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Solution

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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Chapter 2.3: Indefinite Integration - Short Answers II

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