Find the second derivative

[janeann]

d^2/dx^2 Integral from 0 to x^2 Integral from 1 to sin(t) sqrt(1-u^2) du dt.

 

[Deleted member 4993]

d^2/dx^2 Integral from 0 to x^2 Integral from 1 to sin(t) sqrt(1-u^2) du dt.

Have you posted the problem correctly?

You have two variables within the integral - however you have only one set of limit of integration.

 

[janeann]

the problem has d^2/dx^2 then it has two integral signs that are side by side the first one is from 0 to x^2 and the second is from 1 to sin(t). then right next to the second one is sqrt(1-u^2) du dt.

 

[Deleted member 4993]

the problem has d^2/dx^2 then it has two integral signs that are side by side the first one is from 0 to x^2 and the second is from 1 to sin(t). then right next to the second one is sqrt(1-u^2) du dt.

Okay then ... the problem looks like

\(\displaystyle \frac{d^2}{dx^2}\left [\int_0^{x^2}\left (\int_1^{sin(t)}\frac{du}{\sqrt{1-u^2}}\right ) dt \right] \ = \ ????\)

So what is your first step?

 

[roesing]

We were told to then group it so that f(t)=d^2/dx^2 integral from 1 to sin(t) of square root (1-u^2)du and then compute d^2/dx^2 integral from 0 to x^2 of f(t)dt in the abstract then go back and deal with the derivative of f(t) using the fundamental theorem of calculus. I'm not sure how to compute that integral of f(t) in the abstract though

 

[Deleted member 4993]

We were told to then group it so that f(t)=d^2/dx^2 integral from 1 to sin(t) of square root (1-u^2)du and then compute d^2/dx^2 integral from 0 to x^2 of f(t)dt in the abstract then go back and deal with the derivative of f(t) using the fundamental theorem of calculus. I'm not sure how to compute that integral of f(t) in the abstract though

I have grouped it for you - do the integration.

 

[MCROJAS]

How do you start integrating that? Wouldn't it be zero since d/dx of f(variable other than x) = 0?

 

[Deleted member 4993]

How do you start integrating that? Wouldn't it be zero since d/dx of f(variable other than x) = 0?

This is very interesting - 3 different people asking about the same problem - without showing a line of work!!!

\(\displaystyle \frac{d^2}{dx^2}\left [\int_0^{x^2}\left (\int_1^{sin(t)}\frac{du}{\sqrt{1-u^2}}\right ) dt \right]\)

\(\displaystyle = \ \frac{d^2}{dx^2}\left [\int_0^{x^2}\left [ sin^{-1}(u)\right ]_1^{sin(t)} dt \right]\)

\(\displaystyle = \ \frac{d^2}{dx^2}\left [\int_0^{x^2}\left [ t - \frac{\pi}{2}\right ] dt \right]\)

Now continue.....