- #1

The Head

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- Homework Statement
- Evaluate d/dx ∫(limits: 0,x) x*sin(t)

- Relevant Equations
- d/dx ∫ (0,x) f(t)= f(x)*x'

uv-∫vdu = ∫udv

I was told this problem could simply be solved with calc-1 techniques, so I'm tempted to say we could do d/dx(x∫(limits: 0,x) sin(t) dt. Then it's a simple product rule: d/dx (x) * ∫(0,x) sint dt + x * d/dx(∫ (0,x) sin (t) dt) = 1 - cos(x) + x*sin(x). However, I wonder if we have to allow that x is a function of t and then I would be forbidden from bringing x out of the integral. Is this allowed?

I tried moving the d/dx inside the integral too (∫ d/dx (x*sint) dt = ∫ sint dt + ∫x*d/dx(sint) dt.

∫ sint dt = 1 - cos(x)

∫x*d/dx(sint) dt means integration by parts (not really calc 1).

u=x

du=dx/dt

dv= d/dx(sint)

v =∫ d/dx (sint), but now the variables are getting all screwed up because the integral is with respect to t, but I'm taking d/dx. Was my first approach legal enough?

I tried moving the d/dx inside the integral too (∫ d/dx (x*sint) dt = ∫ sint dt + ∫x*d/dx(sint) dt.

∫ sint dt = 1 - cos(x)

∫x*d/dx(sint) dt means integration by parts (not really calc 1).

u=x

du=dx/dt

dv= d/dx(sint)

v =∫ d/dx (sint), but now the variables are getting all screwed up because the integral is with respect to t, but I'm taking d/dx. Was my first approach legal enough?