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anonymous
 4 years ago
how do you derive sin(lnx)dx
anonymous
 4 years ago
how do you derive sin(lnx)dx

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry yes i meant integrate!

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.0let \(u=\sin(\ln x)\) 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

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.0the integral after "Therefore we have" on the left hand side should be \[\int \sin(\ln x) dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let lnx=t then 1/xdx=dt dx=xdt x=e^t dx=e^tdt \[\int\limits e ^{t}sint dt\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sint=u =>costdt=du e^tdt=dv =>e^t=v \[uv\int\limits vdu\] \[e ^{t} sint\int\limits e ^{t}costdt\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do it same thing for cost

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cost=u => sintdt=du e^tdv=dv =>e^t=v \[uv\int\limits\limits vdu\] \[e ^{t} cost+\int\limits e ^{t}sintdt\] \[\int\limits\limits e ^{t}sint dt=e ^{t}sint(e^tcost+\int\limits e^t sint)\] \[2\int\limits\limits\limits e ^{t}sint dt=e ^{t}sinte^tcost\] \[\int\limits\limits\limits\limits e ^{t}sint dt=\frac1{2}(e ^{t}sinte^tcost)\] 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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