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

1. watchmath

you mean integrate?

2. anonymous

cos(lnx)/x

3. anonymous

sorry yes i meant integrate!

4. watchmath

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

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

6. anonymous

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

7. anonymous

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

8. anonymous

do it same thing for cost

9. anonymous

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