# Homework Help: Should be easy integration

1. Jan 31, 2010

### holezch

1. The problem statement, all variables and given/known data
F(x) = sin ($$\int$$0 to x sin ( $$\int$$ 0 to y sin^3(u) du) dy )

find F'(x)
2. Relevant equations
FTC

3. The attempt at a solution

cos ($$\int$$0 to x sin ( $$\int$$ 0 to y sin^3(u) du) dt ) sin( $$\int$$ 0 to x sin^3(u) du) dy ) sin^3(x)?

because when you differentiate the integral, the function you get becomes with respect to the boundary variable and so the last terms become with respect to x and I have to differentiate them as well?

($$\int$$0 to x sin ( $$\int$$ 0 to y sin^3(u) du) dy ) this becomes sin ( tex]\int[/tex] 0 to x sin^3(u) du), then since the integral in the argument is with respect to x, I must differentiate that as well to get sin^3(x)

thanks

Last edited: Jan 31, 2010
2. Jan 31, 2010

### LCKurtz

I doubt anyone can make sense of that. Use a little tex to make it readable. For example, you do an integral like this:

$$\int_{lower limit here}^{upper limit here} integrand\ dx [/itex] To see what I did just click on it. You can copy and paste it and use it to post your question so we can read it. 3. Jan 31, 2010 ### holezch thanks, I didn't know how to do that anyway, I'm looking for the derivative of: [tex] F(x) = sin(\int_{0}^{x} sin(\int_{0}^{y} sin^3(u)\ du)\ dy )$$

I'm pretty sure that it comes out to be:

$$F'(x) = [cos(\int_{0}^{x} sin(\int_{0}^{y} sin^3(u)\ du)\ dy ) \ dx] sin(\int_{0}^{x} sin^3(u)\ du) sin^3(x)$$

But I'm not 100% clear about the steps. Why do I have to keep differentiating after I differentiate $$(\int_{0}^{x} sin(\int_{0}^{y} sin^3(u)\ du)\ dy )$$ (from the chain rule)

Is it because when I differentiate the expression, $$sin(\int_{0}^{y} sin^3(u)\ du)\ dy )$$, it is evaluated at x and so I must differentiate that expression? (since I am differentiating with respect to x)

thank you

4. Jan 31, 2010

### LCKurtz

You have an extra dx at the end of your F(x) that shouldn't be there.

Call

$$g(x)= \int_{0}^{x} sin(\int_{0}^{y} sin^3(u)\ du)\ dy$$

so your F(x) = sin(g(x)) and F'(x) = cos(g(x)) g'(x) by the chain rule.

Now by the fundamental theorem of calculus to get g'(x) you simply write the integrand in g(x) with the dummy variable y replaced by x:

$$g'(x) = sin(\int_{0}^{x} sin^3(u)\ du)$$

Put those together; I think it is a bit different than what you wrote. It's late and I'm a little sleepy so check it. :zzz:

5. Jan 31, 2010

### holezch

thanks, I noticed the extra dx. I left it there by accident after copying and pasting the integral latex code

by the chain rule it should be F'(x) = cos(g(x) )g'(x), but how come the answer has sin^3(x) at the end? is it wrong? (I got this answer from someone's notes, refer to post #3)
thanks

6. Feb 1, 2010

### holezch

forgive me for bumping.. but bump