Question

Integrate sin(logx)

Answer 1

(Indefinite integration)

Answer 2

lmfaoooo @DLS now that you changed the question, perhaps it's more solvable :D

Answer 3

I have options on this one,depends whichever we want to solve :P this is easier

Answer 4

@DLS btw is log the natural logarithm or base 10?

Answer 5

nothing given as such. e i suppose

Answer 6

try using by parts

Answer 7

If this is base 10 then yeah, would have to do by parts I would think. u = sin(logx) and v = dx.

Answer 8

like 1.sin(logx) ?

Answer 9

yes

Answer 10

If 'e' is the base then log is just ln. Now, we start by making a u-substitution:\[\bf u=\ln(x) \implies \frac{ du }{ dx }=\frac{ 1 }{x \ln(e) }=\frac{ 1 }{ x } \implies dx=xdu=e^u \ du\]Now re-write the integral as:\[\bf \int\limits_{}^{}\sin(u)(e^u \ du)=\int\limits_{}^{}e^usin(u) \ du\]At this point it would be wise to use integration by parts which yields:\[\bf \implies \int\limits_{}^{}e^usin(u) \ du =e^usin(u)-\int\limits_{}^{}e^u \cos(u) \ du\]In this case I chose my \(\bf v=e^u\).

Now let's do integration by parts again but this time we'll choose \(\bf v=sin(u)\):\[\bf \implies \int\limits_{}^{}e^usin(u) \ du=-e^ucos(u)+\int\limits_{}^{}e^ucos(u) \ du\]Now the MAGIC happens when you add the two integrals that results from the use of parts. Adding these 2 integrals we get:\[\bf \implies 2\int\limits_{}^{}e^usin(u) \ du=e^usin(u)-e^ucos(u)=e^u[\sin(u)-\cos(u)]\]Now divide both sides by 2:\[\bf \implies \int\limits_{}^{}e^usin(u) \ du=\frac{ e^u }{ 2 }[\sin(u)-\cos(u)]\]Now substitute back in for \(\bf u = ln(x)\) and use the identity \(\bf e^{ln(a)}=a\) to get:\[\bf \implies \int\limits_{}^{}\sin[\ln(x)] \ dx=\frac{ x }{ 2 }[\sin[\ln(x)]-\cos[\ln(x)]]+C\]And we are done :D @DLS

Answer 11

If it's base 10 then I think I got ya on that, haha.

Answer 12

@DLS you afk or something?

Answer 13

thanks! and yes i was away sorry,i got it :)