How Do You Solve the Integral of 5sin(lnx)?

[Drakkith]

Homework Statement

Make a substitution and then evaluate the integral.
∫5 sin(lnx) dx

Homework Equations

The Attempt at a Solution

Let t = lnx
e^t = e^lnx
e^t = x
dt = 1/x dx
dx = xdt
dx = e^t dt

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

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

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

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

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

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.

 

[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 $$\int \sin(\ln(x)) \ dx.$$

 

[SammyS]

Homework Statement

Make a substitution and then evaluate the integral.
∫5 sin(lnx) dx

Homework Equations

The Attempt at a Solution

Let t = lnx
e^t = e^lnx
e^t = x
dt = 1/x dx
dx = xdt
dx = e^t dt

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

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

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

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

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

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.

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

 

[axmls]

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

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.

 

[Drakkith]

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

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

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

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!