Integration of cos(logx)dx?

1 Answer
Apr 20, 2018

intcos(lnx)dx=1/2(xsin(lnx)+xcos(lnx))+C

Explanation:

So, we have

intcos(lnx)dx

We'll make the following substitution:

u=lnx

du=dx/x

xdu=dx

Let's try to get x in terms of u:

e^u=e^lnx

e^u=x

Thus,

e^udu=dx and we get

inte^ucosudu. We now use Integration by Parts twice, making the following selections:

w=e^u
dw=e^udu
dv=cosudu
v=sinu

Applying the formula uw-intvdw:

inte^ucosudu=e^usinu-inte^usinudu

For inte^usinudu:

w=e^u
dw=e^udu
dv=sinudu
v=-cosu

Thus,
inte^usinudu=-e^ucosu+inte^ucosudu

We see our original integral shows back up:

inte^ucosudu=e^usinu-(-e^ucosu+inte^ucosudu)

inte^ucosudu=e^usinu+e^ucosu-inte^ucosu

2inte^ucosudu=e^usinu+e^ucosu

inte^ucosudu=1/2(e^usinu+e^ucosu)+C (Don't forget the constant of integration)

Recalling that u=lnx, e^u=x:

intcos(lnx)dx=1/2(xsin(lnx)+xcos(lnx))+C