# Integral of 5 sin(lnx)

1. Feb 23, 2016

### Drakkith

Staff Emeritus
1. The problem statement, all variables and given/known data
Make a substitution and then evaluate the integral.
∫5 sin(lnx) dx

2. Relevant equations

3. The attempt at a solution
Let t = lnx
et = elnx
et = x
dt = 1/x dx
dx = xdt
dx = et dt

Now the integral is: 5∫sin(t) et dt
Integrating by parts:
u = sin(t), du = cos(t) dt
dv = et dt, v = et
5(sin(t)et - ∫cos(t)et dt)

By parts again:
u=cos(t), du=-sin(t) dt
dv = et dt, v = et
5[sin(t)et - (cos(t)et + ∫sin(t)et dt)]

Distributing the 5 and the negative sign:
5∫sin(t)et dt = 5sin(t)et - 5cos(t)et - 5∫sin(t)et dt]

Bringing the integral over the left:
10∫sin(t)et = 5sin(t)et - 5cos(t)et

Dividing the 10 out:
∫sin(t)et = 1/2(sin(t)et - cos(t)et)

Substituting lnx = t
1/2x[sin(lnx) - cos(lnx)] +C

Now, supposedly the answer is 5/2x [sin(lnx)-cos(lnx)] + C, but I can't figure out why.

2. Feb 23, 2016

### axmls

You shouldn't have divided that 10 out. The antiderivative you wanted to find was $$\int 5 \sin(\ln(x)) \ dx,$$ but you found $$\int \sin(\ln(x)) \ dx.$$

3. Feb 23, 2016

### SammyS

Staff Emeritus
It looks like you may have dropped a sign in that last integration by parts.

4. Feb 23, 2016

### axmls

I don't think so. The sine term is negative, which flips the sign of that second integration by parts integral. If the OP follows my post, the answer matches the one given.

5. Feb 23, 2016

### Drakkith

Staff Emeritus
As axmls said, the sine term turns that final integral positive.

Oh. I had no idea that's how that worked. So there's a 10 on the left, a 5 on the right, and dividing both sides by 2 makes the left side 5 and the right side 5/2, which would be the correct answer.

Thanks guys!