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anonymous

  • 5 years ago

how do you derive sin(lnx)dx

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  1. watchmath
    • 5 years ago
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    you mean integrate?

  2. anonymous
    • 5 years ago
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    cos(lnx)/x

  3. anonymous
    • 5 years ago
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    sorry yes i meant integrate!

  4. watchmath
    • 5 years ago
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    let \(u=\sin(\ln x)\) and \(dv=dx\). Then \(du=\cos(\ln x)/x dx\) and \(v=x\). Using integration by parts we have \[x\sin(\ln x)-\int \cos(\ln x)dx\] Using the same method we can write \[\int \cos(\ln x) dx= x\cos(\ln x)+\int \sin(\ln x)dx\] Therefore we have \[\int \cos(\ln x)dx=xsn(\ln x)-x\cos(\ln x)-\int \sin(\ln x)dx\] Therefore (after combining the two integrals) \[2\int \sin(\ln x)dx=x(\sin(\ln x)-\cos(\ln x))\] Divide bothsides by 2 and we are don

  5. watchmath
    • 5 years ago
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    the integral after "Therefore we have" on the left hand side should be \[\int \sin(\ln x) dx\]

  6. anonymous
    • 5 years ago
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    let lnx=t then 1/xdx=dt dx=xdt x=e^t dx=e^tdt \[\int\limits e ^{t}sint dt\]

  7. anonymous
    • 5 years ago
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    sint=u =>costdt=du e^tdt=dv =>e^t=v \[uv-\int\limits vdu\] \[e ^{t} sint-\int\limits e ^{t}costdt\]

  8. anonymous
    • 5 years ago
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    do it same thing for cost

  9. anonymous
    • 5 years ago
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    cost=u => -sintdt=du e^tdv=dv =>e^t=v \[uv-\int\limits\limits vdu\] \[e ^{t} cost+\int\limits e ^{t}sintdt\] \[\int\limits\limits e ^{t}sint dt=e ^{t}sint-(e^tcost+\int\limits e^t sint)\] \[2\int\limits\limits\limits e ^{t}sint dt=e ^{t}sint-e^tcost\] \[\int\limits\limits\limits\limits e ^{t}sint dt=\frac1{2}(e ^{t}sint-e^tcost)\] lnx=t \[e ^{lnx}=x\] \[\int\limits\limits\limits\limits\limits\limits\limits e ^{t}sint dt=\frac1{2}x\sin(lnx)-\frac1{2}x\cos(lnx)\]

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