- #1

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## 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.