Question

how do you derive sin(lnx)dx

Answer 1

you mean integrate?

Answer 2

cos(lnx)/x

Answer 3

sorry yes i meant integrate!

Answer 4

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

Answer 5

the integral after "Therefore we have" on the left hand side should be
\[\int \sin(\ln x) dx\]

Answer 6

let lnx=t
then 1/xdx=dt
dx=xdt
x=e^t
dx=e^tdt
\[\int\limits e ^{t}sint dt\]

Answer 7

sint=u =>costdt=du
e^tdt=dv =>e^t=v
\[uv-\int\limits vdu\]
\[e ^{t} sint-\int\limits e ^{t}costdt\]

Answer 8

do it same thing for cost

Answer 9

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

\[e ^{lnx}=x\]
\[\int\limits\limits\limits\limits\limits\limits\limits e ^{t}sint dt=\frac1{2}x\sin(lnx)-\frac1{2}x\cos(lnx)\]